a: (2x-3)(3x+6)>0

=>(2x-3)(x+2)>0

=>x<-2 hoặc x>3/2

b: (3x+4)(2x-6)<0

=>(3x+4)(x-3)<0

=>-4/3<x<3

c: (3x+5)(2x+4)>4

\(\Leftrightarrow6x^2+12x+10x+20-4>0\)

\(\Leftrightarrow6x^2+22x+16>0\)

=>\(6x^2+6x+16x+16>0\)

=>(x+1)(3x+8)>0

=>x>-1 hoặc x<-8/3

f: (4x-8)(2x+5)<0

=>(x-2)(2x+5)<0

=>-5/2<x<2

h: (3x-7)(x+1)<=0

=>x+1>=0 và 3x-7<=0

=>-1<=x<=7/3

ảnh ko theo trật tự và bị thiếu nên mk sẽ gửi lại 1 tấm nx và mong bn thông cảm cho ​

a) Ta có: \(\left(2x-8\right)\left(2x+10\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8\ge0\\2x+10\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-5\end{matrix}\right.\)

b) Ta có: \(\left(\left|x\right|+5\right)\left(x-3\right)< 0\)

nên x-3<0

hay x<3

a, (5x+7)(2x-1) <0 

<=> \(\hept{\begin{cases}5x+7< 0\\2x-1>0\end{cases}}\)<=> \(\hept{\begin{cases}5x< 7\\2x< 1\end{cases}}\)

<=> \(\hept{\begin{cases}5x+7>0\\2x-1< 0\end{cases}}\)<=> ..................

(5x+7)(2x-1) =0 

<=> \(\orbr{\begin{cases}5x+7=0\\2x-1=0\end{cases}}\)<=> ..................

a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)

=>\(\left|2x+1\right|=\dfrac{1}{2}\)

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

c: (2x-3)²=36

=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

d: \(7^{x+2}+2\cdot7^x=357\)

=>\(7^x\cdot49+7^x\cdot2=357\)

=>\(7^x=7\)

=>x=1

a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(---\)

b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)

\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)

\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

\(---\)

c) \(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(---\)

d) \(7^{x+2}+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)

\(\Rightarrow7^x\cdot\left(49+2\right)=357\)

\(\Rightarrow7^x\cdot51=357\)

\(\Rightarrow7^x=357:51\)

\(\Rightarrow7^x=7\)

\(\Rightarrow x=1\)

a)(2x-3)²=1<=> \(\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}< =>\orbr{\begin{cases}2x=4\\2x=2\end{cases}}}\)\(< =>\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

x=2 =>2².5²=20^y.5^y <=>100 = 100^y <=> y=1

x=1 => 2.5= 20^y.5^y <=>10=100^y <=>y = 1/2

b)(4x-3)²+(y²-9)²\(\ge0\)

dấu = sảy ra khi \(\hept{\begin{cases}4x-3=0\\y^2-9=0\end{cases}< =>\hept{\begin{cases}4x=3\\y^2=9\end{cases}}}\)\(\hept{\begin{cases}x=\frac{3}{4}\\y=\pm3\end{cases}}\)

c) <=> (y-5)⁸ \(\le-\left(x+4\right)^7\)     (1)

(y-5)⁸ >=0 với mọi y nên -(x+4)⁷ \(\ge\left(y-5\right)^8\ge0\)<=> (x+4)⁷\(\le0< =>x+4\le0< =>x\le-4\)

Khi đó (1) <=> y-5\(\le\sqrt[8]{-\left(x+4\right)^7}\) <=> y\(\hept{\begin{cases}y\le5-\sqrt[8]{-\left(x+4\right)^7}\\x\le-4\end{cases}}\) 

Em hc bảng xét dáu chx ??

Lớp 7 chưa học đâu em

4: \(\left|x^3-64\right|+\left|15-4y\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-64=0\\15-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{15}{4}\end{matrix}\right.\)

6: |7x-11|>5

=>7x-11>5 hoặc 7x-11<-5

=>7x>16 hoặc 7x<6

=>x>16/7 hoặc x<6/7

8: |2x+12|<4

=>2x+12>-4 và 2x+12<4

=>2x>-16 và 2x<-8

=>-8<x<-4