a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)

c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)

=>x+3=0 hoặc x-4=0

=>x=-3 hoặc x=4

e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)

f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)

hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)

a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)

b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)

c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)

d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

e, tương tự d 

f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)

bạn tải ảnh về r up lại đi bạn

\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)

\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)

\(\Leftrightarrow-28x+37\ge12\)

\(\Leftrightarrow-28x\ge12-37\)

\(\Leftrightarrow-28x\ge-25\)

\(\Leftrightarrow x\le\dfrac{25}{28}\)

Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)

b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)

\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)

\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)

\(\Leftrightarrow-6x\ge30\)

\(\Leftrightarrow x\le-5\)

Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)

\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)

\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)

\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)

\(\Leftrightarrow-11x+37< 0\)

\(\Leftrightarrow-11x< -37\)

\(\Leftrightarrow x>\dfrac{37}{11}\)

vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)

bài này bạn xét dấu gtri tuyệt đối ý

|x-4|+|x-9|=|x-4|+|9-x| >/ |x-4+9-x|=5

đẳng thức xảy ra <=> (x-4)(9-x) >/ 0 <=> 4 </ x </ 9

đến đây tự kết luân

a) \(\left|4+2x\right|=-4x\)

TH1 : \(4+2x\ge0\Leftrightarrow2x\ge-4\Leftrightarrow x\ge-2\)

\(4+2x=-4x\)

\(\Leftrightarrow2x+4x=-4\)

\(\Leftrightarrow6x=-4\)

\(\Leftrightarrow x=-\dfrac{2}{3}\) (t/m)

TH2 : \(4+2x< 0\Leftrightarrow2x< -4\Leftrightarrow x< -2\)

\(\text{- (4 + 2x) = -4x}\)

\(\Leftrightarrow-4-2x=-4x\)

\(\Leftrightarrow-2x+4x=4\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\) (ko t/m)

\(S=\left\{-\dfrac{2}{3}\right\}\)

b) \(\left|-2,5x\right|=x-12\)

TH1 : \(-2,5x\ge0\Leftrightarrow x\le0\)

\(-2,5x=x-12\)

\(\Leftrightarrow-2,5x-x=-12\)

\(\Leftrightarrow-3,5x=-12\)

\(\Leftrightarrow x=\dfrac{24}{7}\) (ko t/m)

TH2 : \(-2,5x< 0\Leftrightarrow x>0\)

\(\text{2,5x = x - 12}\)

\(\Leftrightarrow2,5x-x=-12\)

\(\Leftrightarrow1,5x=-12\)

\(\Leftrightarrow x=-8\) (ko t/m)

\(S=\varnothing\)

c) \(\left|-2x\right|+x-5x-3=0\)

\(\Leftrightarrow\left|-2x\right|-4x-3=0\)

\(\Leftrightarrow\left|-2x\right|=3+4x\)

TH1 : \(-2x\ge0\Leftrightarrow x\le0\)

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

\(\Leftrightarrow-2x-4x=3\)

\(\Leftrightarrow-6x=3\)

\(\Leftrightarrow x=-\dfrac{1}{2}\) (t/m)

TH2 : \(-2x< 0\Leftrightarrow x>0\)

\(\text{2x = 3 + 4x}\)

\(\Leftrightarrow2x-4x=3\)

\(\Leftrightarrow-2x=3\)

\(\Leftrightarrow x=-\dfrac{3}{2}\) (ko t/m)

\(S=\left\{-\dfrac{1}{2}\right\}\)

Đặt:\(\sqrt{x+9}=v;\sqrt{x+6}=u\)

Ta có: \(v+5u=5+vu\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\Leftrightarrow\left(5-v\right)\left(u-1\right)\)

\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(N\right)\end{matrix}\right.ĐKXĐ:x>=-6\)

\(S=\left\{16,-5\right\}\)

Câu trên mình quên -5>-6

Đặt: \(\sqrt{x+9}=v;\sqrt{x+6}=u\)

Ta có: \(v+5u=5+vu\)

\(\Leftrightarrow v+5u-5-uv=0\)

\(\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\)

\(\Leftrightarrow\left(5-v\right)\left(u-1\right)\)

\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(L\right)\end{matrix}\right.\)          ĐKXĐ:\(x>=-6\)

\(S=\left\{16\right\}\)

\(\sqrt{x+9}+5\sqrt{x+\text{6}}=5+\sqrt{\left(x+9\right)\left(x+\text{6}\right)}\Leftrightarrow\sqrt{x+9}+5=5+\sqrt{x+9}\Leftrightarrow\sqrt{x+9}-\sqrt{x+9}=0\Leftrightarrow x+9-x-9=0\Leftrightarrow0=0\)

Vậy x vô số nghiệm

Ta có: \(\left\{{}\begin{matrix}3\left|x-1\right|+2\left(x-y\right)=4\\4\left|x-1\right|-\left(x-y\right)=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12\left|x-1\right|+8\left(x-y\right)=16\\12\left|x-1\right|-3\left(x-y\right)=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\left(x-y\right)=-11\\3\left|x-1\right|+2\left(x-y\right)=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\3\left|x-1\right|=4-2\left(x-y\right)=4-2\cdot\left(-1\right)=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\\left|x-1\right|=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y=-1\\x-1=2\end{matrix}\right.\\\left\{{}\begin{matrix}x-y=-1\\x-1=-2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x+1=3+1=4\\x=3\end{matrix}\right.\\\left\{{}\begin{matrix}y=x+1=-1+1=0\\x=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(3;4\right);\left(-1;0\right)\right\}\)

À thôi mik tự làm đc rồi ạ !