a) \(\left(4x-1\right)^2-\left(x+2\right)^2=0\)

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

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

\(\Leftrightarrow3\left(5x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x+1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\x=1\end{cases}}\)

b) \(x^2-7x=8\Leftrightarrow x^2-7x-8=0\)

\(\Leftrightarrow x^2+x-8x-8=0\)

\(\Leftrightarrow x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=8\end{cases}}\)

c) \(\left(5x-7\right)^2-25=0\Leftrightarrow\left(5x-7-5\right)\left(5x-7+5\right)=0\)

\(\Leftrightarrow\left(5x-12\right)\left(5x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-12=0\\5x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{5}\\x=\frac{2}{5}\end{cases}}\)

A=(x−1)2+8≥8Amin=8⇔x=1B=(x+3)2−12≥−12Bmin=−12⇔x=−3C=x2−4x+3+9=(x−2)2+8≥8Cmin=8⇔x=2E=−(x+2)2+11≤11Emax=11⇔x=−2F=9−4x2≤9Fmax=9⇔x=0

HT

A=x²-2x+9

Ta có: A=x^2-2x+9

=> A=(x^2-2x+1)+8

=>A=(x-1)^2+8

vì (x-1)^2 > 0 với mọi x

=> (x-1)^2+8> 8 với mọi x

Dấu "=" xáy ra khi:

 (x-1)^2=0=>x-1=0=>x=0+1=>x=1

Vậy Amin = 8 khi x=1

B=x^2+6x-3

=>B=-(x^2-6x+3)

=>B=-(x^2-2.3x+3^2)-3

=>B=-(x-3)^2-3

vì -(x-3)^2 < 0 với mọi x

=>-(x-3)^2-3< -3 với mọi x

Dấu '=' xảy ra khi x-3=0=>x=0+3=>x=3

Vậy B(min)=-3 khi x=3

chỗ này hình như là Bmax xem lại đề nhé

D=-x^2-4x+7

=>D=-x^2-2.2x+4+3

=>D=(-x^2-2.2x+4)+3

=>D=(-x-2)^2+3

Vì (-x-2)^2 <0 với mọi x

=>(-x-2)^2+3<3 với mọi x

Dấu "=" xảy ra khi x-2=0=>x=0+2=>x=2

Vậy Dmax=3 khi x=2

E=5-4x^2+4x

=>E=-4x^2+4x+5

=>E=(-2x)^2+2.2x+4+1

=>E=[(-2x)^2+2.2x+4]

=>E=(-2x+2)^2+1

Vì: (-2x+2)^2 < 0 với mọi x

=>(-2x+2)^2+1 <  1 với mọi x

Dấu "=" xảy ra khi 2x+2=0=>2x=-2=>x=-1

Vậy Emax=1 khi x=-1

Bài 209 : đăng tách ra cho mn cùng làm nhé 

a,sửa đề :  \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)

b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)

\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)

c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)

\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)

\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)

\(1,4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

\(4x^2-20x-4x^2+3x-4x+3=5\)

\(-21x+3=5\)

\(21x=-8\)

\(x=-\frac{8}{21}\)

\(2,8x^3-50x=0\)

\(x\left(8x^2-50\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\8x^2-50=0\Rightarrow x=\pm2\end{cases}}\)

Vậy ....

\(3,\left(2x-1\right)^2-25=0\)

\(\left(2x-1\right)^2=\pm5^2\)

\(\Rightarrow\hept{\begin{cases}2x-1=5\\2x-1=\left(-5\right)\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=\left(-2\right)\end{cases}}\)

Vậy ... 

Bài 1:a. (x+3)2−(x−4)(x+8)=1⇔x2+6x+9−x2−4x+32=1⇔2x=−40⇔x=20Vậy S={20}b. 4x−20+3x−15=0⇔7x=35⇔x=5Vậy S={5}c. x3−5x2+25x+5x2−25x+125−x3=5x⇔5x=125⇔x=25Vậy S={25}d. 4x2+4x+1−4x2+9=22⇔4x=12⇔x=3Vậy S={3}e. 3x−3−1+x=0⇔4x=4⇔x=1Vậy S={1}f. x2(x+3)−5(x+3)=0⇔(x+3)(x2−5)=0⇔x=−3; x=±√5Vậy S={−3; ±√5}Bài 2:a. x2−6x+9=0⇔(x−3)2=0⇔x=3Vậy S={3}b. 8x3−12x2+6x−1=0⇔(2x−1)3=0⇔x=12Vậy S={12}c. x3+4x2+4x=0⇔x(x2+4x+4)=0⇔x(x+2)2=0⇔x=0; x=−2Vậy S={−2;0}d. 4x3−36x=0⇔4x(x2−9)=0⇔x=0; x=±3Vậy S={0;±3}e. x3+5x2−4x−20=0⇔(x−2)(x+2)(x+5)=0⇔x=±2; x=−5Vậy S={±2;−5}f. 2x2+16x+32−x2+4=0⇔x2+16x+36=0⇔(x+8)2=28⇔x=−8±2√7Vậy S={−8±2√7}g.x3−27+4x−x3=0⇔4x=27⇔x=274Vậy S={274}h. x2+5x−14=0⇔(x−2)(x+7)=0⇔x=2; x=−7Vậy  S={2;−7}

\(4x^2-25+\left(2x+7\right).\left(5-2x\right)\)

\(=\left(2x+5\right).\left(2x-5\right)-\left(2x+7\right).\left(2x-5\right)\)

\(=\left(2x+5-2x-7\right).\left(2x-5\right)\)

\(=-2.\left(2x-5\right)\)

\(a^2x^2-a^2x^2-b^2x^2+b^2y^2\)

\(=a^2.\left(x^2-y^2\right)-b^2.\left(x^2-y^2\right)\)

\(=\left(a^2-b^2\right).\left(x^2-y^2\right)\)

\(=\left(a-b\right).\left(a+b\right).\left(x-y\right).\left(x+y\right)\)

\(x^2-y^2+12y-36\)

\(=x^2-\left(y^2-12y+36\right)\)

\(=x^2-\left(y-6\right)^2\)

\(=\left(x-y+6\right).\left(x+y-6\right)\)

\(\left(x+2\right)^2-x^2+2x-1\)

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

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

\(=[x+2-\left(x-1\right)].[x+2+\left(x-1\right)]\)

\(=\left(x+2-x+1\right).\left(x+2+x-1\right)\)

\(=3.\left(2x+1\right)\)

\(16x^2-y^2=\left(4x\right)^2-y^2=\left(4x-y\right).\left(4x+y\right)\)

\(1+27x^3=1^3+\left(3x\right)^3=\left(1+3x\right).\left(1-3x+9x^2\right)\)