2 (3x -1) (2x +5) - 6 (2x - 1) (x + 2) = -6

\(\Leftrightarrow2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)

\(\Leftrightarrow12x^2+30x-4x-10-12x^2-24x+6x+12=-6\)

\(\Leftrightarrow8x=-8\Leftrightarrow x=-1\)

Vậy x = -1

xin lỗi bạn 

mình chưa học cái này nên mình không biết

Ta có: \(2n^3-7n^2+2n+12=\left(2n+3\right)\left(n^2-5n+8\right)+n-12\)

chia hết cho \(2n+3\)tương đương với \(\left(n-12\right)⋮\left(2n+3\right)\)

\(\Leftrightarrow2\left(n-12\right)=2n-24=2n+3-27⋮\left(2n+3\right)\Leftrightarrow27⋮\left(2n+3\right)\)

mà \(n\)nguyên nên \(2n+3\inƯ\left(27\right)=\left\{-27,-9,-3,-1,1,3,9,27\right\}\)

\(\Leftrightarrow n\in\left\{-15,-6,-3,-2,-1,0,3,12\right\}\).

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

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

\(-21x+3=5\)

\(21x=-8\)

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

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

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

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

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}

LOVE YOU BABY

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Gửi bạn nè

ta cos :

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

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

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

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

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

Gửi bạn nè. Chúc bạn học tốt !