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Bài 1: Giải phương trình
a) ĐKXĐ: \(x\ge3\)
Ta có: \(\sqrt{100\cdot\left(x-3\right)}=\sqrt{20}\)
\(\Leftrightarrow\left|100\cdot\left(x-3\right)\right|=\left|20\right|\)
\(\Leftrightarrow100\cdot\left|x-3\right|=20\)
\(\Leftrightarrow\left|x-3\right|=\frac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\frac{1}{5}\\x-3=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{16}{5}\left(nhận\right)\\x=\frac{14}{5}\left(loại\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{16}{5}\right\}\)
b) Ta có: \(\sqrt{\left(x-3\right)^2}=7\)
\(\Leftrightarrow\left|x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
Vậy: S={10;-4}
c) Ta có: \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)
1)\(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow\hept{\begin{cases}7\ge0\\x-1=49\end{cases}\Leftrightarrow x=50}\)
6.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2\end{matrix}\right.\)
4.
ĐKXĐ: \(x\ge4\)
Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)
\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)
\(\Leftrightarrow3t^2-7t+34=0\)
Phương trình vô nghiệm
5.
ĐKXĐ: ...
- Với \(x=0\) ko phải nghiệm
- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:
\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
1. \(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|=1\)
+ Ta có : \(\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|\ge\left|\sqrt{x}-2+3-\sqrt{x}\right|=1\)
Dấu "=" \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
\(\Leftrightarrow2\le\sqrt{x}\le3\Leftrightarrow4\le x\le9\)
2. + \(ĐK:4-2x-x^2\ge0\)
+ VT = \(\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}\)
\(=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\) \(\ge\sqrt{4}+\sqrt{9}=5\) (1)
Dấu "=" \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
+ VP \(=-\left(x^2+2x+1\right)+5=-\left(x+1\right)^2+5\le5\forall x\) (2)
Dấu "=" \(\Leftrightarrow x=-1\)
+ Từ (1) và (2) suy ra : pt \(\Leftrightarrow VT=VP=5\Leftrightarrow x=-1\) (TM)
3. + TH1: \(x< 0\) ta có :
\(VT< \sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )
+ TH2 : x = 0 ta có :
\(VT=\sqrt[3]{1}+\sqrt[3]{0}=1\) ( TM )
+ TH3 : x > 0 ta có :
\(VT>\sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )
Vậy x = 0 là nghiệm duy nhất của pt
4. \(\Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+3\right)-24=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x-8\right)-24=0\)
\(\Leftrightarrow t\left(t-5\right)-24=0\) ( với \(t=x^2+2x-3\) )
\(\Leftrightarrow t^2-5t-24=0\Leftrightarrow\left(t+3\right)\left(t-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-3\\t=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=-3\\x^2+2x-3=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\\left(x+1\right)^2=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=2\sqrt{3}-1\\x=-2\sqrt{3}-1\end{matrix}\right.\) ( TM )
\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{\left(\sqrt{8}-\sqrt{3}\right)\left(\sqrt{8}+\sqrt{3}\right)}\)
\(=\sqrt{3}+1+\sqrt{3}-1+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{5}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{5}\)
\(=2\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{8}+\sqrt{3}\)
\(=4\sqrt{3}\)
Giải pt:
1/ \(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\Rightarrow x=3\)
2/ \(\Leftrightarrow\sqrt{3}x^2=\sqrt{12}\Leftrightarrow x^2=\sqrt{4}=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
3/ \(\Leftrightarrow x-5=9\Rightarrow x=14\)
4/ Đề thiếu
5/ \(\Leftrightarrow\left|x-3\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
6/ \(\Leftrightarrow2\left|1-x\right|=6\)
\(\Leftrightarrow\left|1-x\right|=3\Leftrightarrow\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
7/ \(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow x-1=49\Rightarrow x=50\)
8/ \(\Leftrightarrow x+1=2^3=8\)
\(\Rightarrow x=7\)
9/ \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)
10/ \(\Leftrightarrow\sqrt{2}x=\sqrt{50}\Leftrightarrow x=\sqrt{25}\Rightarrow x=5\)
11/ \(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
12/ \(\Leftrightarrow3-2x=\left(-2\right)^3=-8\)
\(\Leftrightarrow2x=11\Rightarrow x=\frac{11}{2}\)
a) \(13-\sqrt{\left(8x-1\right)^2}=\sqrt{x^2}\) (*)
\(\Leftrightarrow13-\left|8x-1\right|=\left|x\right|\)
Th1: \(8x-1\ge0\Leftrightarrow x\ge\dfrac{1}{8}\)
(*) \(\Leftrightarrow13-8x+1=x\Leftrightarrow9x=14\Leftrightarrow x=\dfrac{14}{9}\left(N\right)\)
Th2: \(x\le0\)
(*) \(\Leftrightarrow13+8x-1=-x\Leftrightarrow9x=-12\Leftrightarrow x=-\dfrac{4}{3}\left(N\right)\)
Th3: \(\left\{{}\begin{matrix}8x-1\ge0\\x\le0\end{matrix}\right.\Leftrightarrow\dfrac{1}{8}\le x\le0\) (vô lý)
Th4: \(\left\{{}\begin{matrix}8x-1\le0\\x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le\dfrac{1}{8}\)
(*) \(\Leftrightarrow13-8x+1=x\Leftrightarrow9x=14\Leftrightarrow x=\dfrac{14}{9}\left(L\right)\)
Kl: x= 14/9 , x= -4/3
b) \(\sqrt{\left(x+1\right)^2}+\sqrt{\left(2x+3\right)^2}=3\Leftrightarrow\left|x+1\right|+\left|2x+3\right|=3\)(*)
Th1: \(x\ge-1\)
(*) \(\Leftrightarrow x+1+2x+3=3\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\left(N\right)\)
Th2: \(x\le-\dfrac{3}{2}\)
(*) \(\Leftrightarrow-x-1-2x-3=3\Leftrightarrow-3x=7\Leftrightarrow x=-\dfrac{7}{3}\left(N\right)\)
Th3: \(\left\{{}\begin{matrix}x+1\ge0\\2x+3\le0\end{matrix}\right.\Leftrightarrow-1\le x\le-\dfrac{3}{2}\) (vô lý)
Th4: \(\left\{{}\begin{matrix}x+1\le0\\2x+3\ge0\end{matrix}\right.\Leftrightarrow-\dfrac{3}{2}\le x\le-1\)
(*) \(\Leftrightarrow-x-1-2x-3=3\Leftrightarrow-3x=7\Leftrightarrow x=-\dfrac{7}{3}\left(L\right)\)
Kl: x= -1/3 , x= -7/3
vt sai chính tả nên thông cảm
cái 2 căn bậc x+1 là căn bậc 2 hay nhân 2