\(B=\frac12\) khi \(x=2\)

a) Xét △MNK và △MNP ta có:

Góc PMN = MKN = 90°

Góc N chung

⇒△MNK ∼ △MNP ( g.g ) (1)

b) Xét △MPK và △NPM có:

\(\) Góc PMN = PKM = 90°

Góc P chung

⇒ △MPK ∼ △MNP ( g.g ) (2)

Từ (1) và (2) suy ra △MNK ∼ △MPK ( g.g )

⇒ \(\frac{NK}{MK}=\frac{MK}{PK}\)

⇒ \(MK^2=NK.PK\)

c) \(MK=\sqrt{KN.KP}=\sqrt{9.4}=6\)

⇒\(Smnp=\frac12.NP.MK=39\left(\operatorname{cm}^2\right)\)

a)\(A=\frac{x^2-2x+1}{x^2-1}=\frac{x-1}{x+1}\)

b) Khi \(x=3\)

\(\frac{3-1}{3+1}=\frac12\)

Khi \(x=-\frac32\)

\(\frac{-\frac32-1}{-\frac32+1}=5\)

c)\(x\in\left(\right.1;-1;2;-2)\)

a)\(7x+2=0\)

\(7x=-2\)

\(x=-\frac27\)

b)\(18-5x=7+3x\) \(\)

\(-8x=-11\)

\(x=\frac{11}{8}\)

225cm

\(a)x-3=(3-x)^2\)

\(\left(\right.x-3)-(3-x)^2=0\)

\(\left(\right.x-3)(3-x-1)=0\)

\(\rArr\left(\right.x-3)=0\) hoặc \((3-x-1)=0\)

\(\rArr x=3\) hoặc \(x=2\)

\(b)x^3+\frac32x^2+\frac34x+\frac18=\frac{1}{64}\)

\(\left(x+\frac12\right)^3=\left(\frac14\right)^3\)

\(x+\frac12=\frac14\)

\(\rArr x=-\frac14\)

\(a)x^2+2xy+y^2-x-y=\left(x^2+2xy+y^2)-\left(x+y\right)\right.=\left(x+y\right)^2-\left(x+y\right)=\left(x+y\right)\left(x+y-1\right)\)

\(b)2x^3+6x^2+12x+8=2\left(x^3+3x^2+6x=4\right)=2\left(x^3+x^2+2x^2+2x+4x+4\right)=2\left\lbrack\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)\right\rbrack=2\left\lbrack x^2\left(x^{}+1\right)+2x\left(x^{}+1\right)+4\left(x+1\right)\right\rbrack=2\left(x+1\right)\left(x^2+2x+4\right)=\)

\(a)x^2+2xy+y^2-x-y=\left(x^2+2xy+y^2)-\left(x+y\right)\right.=\left(x+y\right)^2-\left(x+y\right)=\left(x+y\right)\left(x+y-1\right)\)

\(b)2x^3+6x^2+12x+8=2\left(x^3+3x^2+6x=4\right)=2\left(x^3+x^2+2x^2+2x+4x+4\right)=2\left\lbrack\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)\right\rbrack=2\left\lbrack x^2\left(x^{}+1\right)+2x\left(x^{}+1\right)+4\left(x+1\right)\right\rbrack=2\left(x+1\right)\left(x^2+2x+4\right)=\)

a) Trong △ADC có MP // DC. Áp dụng định lí Thales. Ta có:

\(\frac{AM}{MD}=\frac{AP}{PC}\)

Trong △ABC có PN // AB. Áp dụng định lí Thales. Ta có:

\(\frac{BN}{NC}=\frac{AP}{PC}\)

Từ hai tỉ lệ thức trên, suy ra:

\(\frac{AM}{MD}=\frac{BN}{NC}\)

b) \(MP=2\operatorname{cm}\)

\(PN=\frac43cm\)

\(MN=\frac{10}{3}cm\)

a) Ta có:

\(3x(x−1)−1+x=0\)

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

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

\(\rArr\left(x-1\right)=0\) hoặc \(\left(3x+1\right)=0\)

\(\rArr x=1\) hoặc \(x=-\frac13\)

b) Ta có:

\(x^2-9x=0\)

\(x\left(x-9\right)=0\)

\(\rArr x=0\) hoặc \(\left(x-9\right)=0\)

\(\rArr x=0\) hoặc \(x=9\)