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a. có vấn đề

b.

\(\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-30\)

\(\Leftrightarrow-22x+5x< -30-5\)

\(\Leftrightarrow-17x< -35\)

\(\Leftrightarrow x>\dfrac{35}{17}\)

a) \(8x^2+27=\left(x-1\right)^3+\left(x+4\right)^3\)

\(\Leftrightarrow8x^3+27=x^3-2x^2+x-x^2+2x-1+x^3+8x^2+16x+4x^2+32x+64\)

\(\Leftrightarrow8x^3+27=2x^3+9x^2+51x+63\)

\(\Leftrightarrow8x^3+27-2x^3-9x^2-51x-63=0\)

\(\Leftrightarrow6x^3-36-9x^2-51x=0\)

\(\Leftrightarrow3\left(2x^3-12-3x^2-17x\right)=0\)

\(\Leftrightarrow3\left(2x^2+3x-8x-12\right)\left(x+1\right)=0\)

\(\Leftrightarrow3\left(2x^2+3x-8x-12\right)\left(x+1\right)=0\)

\(\Leftrightarrow3\left[x\left(2x+3\right)-4\left(2x+3\right)\right]\left(x+1\right)=0\)

\(\Leftrightarrow3\left(2x+3\right)\left(x-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x+3=0\\x-4=0\\x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=4\\x=-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=4\\x=-1\end{cases}}\)

tớ tưởng áp dụng công thức: \(\left(A+B\right)^3=A^3+B^3+3AB\left(A+B\right)\)

và \(\left(A-B\right)^3=A^3-B^3-3AB\left(A-B\right)\)

1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow95x-5=96-6x\)

\(\Leftrightarrow95x+6x=96+5\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) 

\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9=32x+60\)

\(\Leftrightarrow30x-32x=60-9\)

\(\Leftrightarrow-2x=51\)

\(\Leftrightarrow x=-\frac{51}{2}\)

3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)

\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow2x+1=5x+1\)

\(\Leftrightarrow2x=5x\)

\(\Leftrightarrow x=0\)

4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)

=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)

=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)

=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)

=> 27 - 9x + 80 - 16x = 12 - 12x - 48

=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0

=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0

=> 143 - 13x = 0

=> 13x = 143

=> x = 11

5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)

=> 6x - 18 + 7x - 35 - 13x - 4 = 0

=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0

=> -57 = 0(vô nghiệm)

6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)

=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)

=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)

=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)

=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)

=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)

=> \(12x+10-10x-3=12x+2\)

=> \(2x+10-3=12x+2\)

=> 2x + 10 - 3 - 12x - 2 = 0

=> (2x - 12x) + (10 - 3 - 2) = 0

=> -10x + 5 = 0

=> -10x = -5

=> x = 1/2

7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)

=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)

=> 6x - 3 - 5x + 10 - x - 7 = 0

=> (6x - 5x - x) + (-3 + 10 - 7) = 0

=> 0x + 0 = 0

=> 0x = 0

=> x tùy ý

Bài 8 tự làm nhé

a: \(\Leftrightarrow4\left(6-x\right)-3x=6\left(2x+3\right)-12\)

=>24-4x-3x=12x+18-12

=>12x+6=-7x+24

=>19x=18

=>x=18/19

b: \(\Leftrightarrow-210x-6\left(x-3\right)-15x=30x+10\left(2x+1\right)\)

=>-225x-6x+18=30x+20x+10

=>-231x+18-50x-10=0

=>-281x=-8

=>x=8/281

c: \(\Leftrightarrow36-2\left(x+3\right)=-4x+1-x\)

=>36-2x-6=-5x+1

=>3x=1+6-36=5-36=-31

=>x=-31/3

d: \(\Leftrightarrow-30\left(x-3\right)+10\left(2x-7\right)=6\left(6-x\right)\)

=>-30x+90+20x-70=36-6x

=>-10x+20=36-6x

=>-4x=16

=>x=-4

d, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow x+10=0\) (Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\) ≠ 0)

\(\Leftrightarrow x=-10\)

Vậy x = -10 là nghiệm của phương trình.

hnuji9on ui bm, 76tfv45tj,

Câu hỏi của Marilyna - Toán lớp 7 | Học trực tuyến

\(a,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[x^2-2x-3-x^2+3x-10\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases};x-13=0}\)

\(\Leftrightarrow x=1;x=2\)hoặc \(x=13\)

\(b,\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)+6\left(x^2+x\right)-12=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)+6\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+2x-x-2\right)=0\)

\(\Leftrightarrow\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)=0\)

Lại do \(x^2+x+6=\left(x+\frac{1}{2}\right)^2+5\frac{3}{4}\ge5\frac{3}{4}>0\)

\(\Rightarrow\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)

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Câu 1 đến câu 5 anh trả lời rồi vào đó xem

Câu 6:

\(-1+b=0\Rightarrow b=1\). Chọn A

1. D

2. B

3. B

4. A

5. B

6. A