Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$

$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$

$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$

$\Leftrightarrow 13x+15=2$

$\Leftrightarrow 13x=2-15=-13$

$\Leftrightarrow x=-13:13=-1$

Bài 2:

$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:

$(y+4)y=5$

$\Leftrightarrow y^2+4y-5=0$

$\Leftrightarrow (y-1)(y+5)=0$

$\Leftrightarrow y=1$ hoặc $y=-5$

Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$

Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

yggucbsgfuyvfbsudy

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Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)

\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)

bài 1:
a 2x(x-5)-2x^2=20
<=>2x^2-10x-2x^2=20
<=>-10x=20
<=>x=-2
v....
b x^2-2x+1=0
<=>(x-1)^2=0
<=>x-1=0
<=>x=1
v...
bài 3

A=x-x^2+1=-(x^2-x-1)=-(x^2-2*x*1/2+1/4-5/4)=-(x-1/2)^2+5/4<=5/4
dấu bằng xảy ra <=>x=1/2
bài 2 mình ko biết làm sorry cậu

tran thu phuong cảm ơn bn nhá.

Ai giúp tớ câu 2 đi

Giúp mình vs.

\(\left|x-1\right|=2x-1\)

\(\Rightarrow\orbr{\begin{cases}x-1=2x-1\\x-1=-2x+1\end{cases}\left(đk:2x\ge\frac{1}{2}\right)}\)

\(\Rightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{2}{3}\left(tm\right)\end{cases}}\)

vậy \(x=\frac{2}{3}\)