ai giúp minh đi cần quá gấp

\(\left(3.x^2-2^4\right)2^3=2^8\\ \left(3.x^2-2^4\right)=2^8:2^3\\ \left(3.x^2-2^4\right)=2^5\\ 3.x^2-2^4=32\\ 3.x^2=32+2^4\\ 3.x^2=48\\ x^2=48:3\\ x^2=16\\ x=+-4\)

bn ghi lại đề câu b) nhé

lx-5l-2(-3)=2^4

lx-5l+6=16

lx-5l=16-6

lx-5l=10

\(\left[{}\begin{matrix}x-5=-10\\x-5=10\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-10+5\\x=10+5\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-5\\x=15\end{matrix}\right.\)

a) ( x + 2 )( x + 3 ) - ( x - 2 )( x + 5 ) = 16

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 16

<=> x² + 5x + 6 - x² - 3x + 10 = 16

<=> 2x + 16 = 16

<=> 2x = 0

<=> x = 0

b) 3x( 2x - 4 ) - 2x( 3x + 5 ) = 44

<=> 6x² - 12x - 6x² - 10x = 44

<=> -22x = 44

<=> x = -2

c) 2( 5x - 8 - 3 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 5x - 11 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 20x² - 69x + 55 ) = 12x - 16

<=> 40x² - 138x + 110 = 12x - 16

<=> 40x² - 138x + 110 - 12x + 16 = 0

<=> 40x² - 150 + 126 = 0 ( chưa học nghiệm vô tỉ nên để vô nghiệm nha :) )

=> Vô nghiệm

a. (x+2) (x+3)-(x-2) (x+5)= 16

x²+5x+6-x²-3x+10=16

2x+16=16

2x=0

x=0

b,3x (2x-4)-2x (3x+5)= 44

6x²-12x-6x²-10x=44

-22x=44

x=-2

Ý c bạn tự lm,tương tự nhưa,b

a, \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=16\)

\(\Leftrightarrow x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=16\)

\(\Leftrightarrow8x-4=16\Leftrightarrow x=\frac{20}{8}=\frac{5}{2}\)

b, \(3x\left(2x-4\right)-2x\left(3x+5\right)=44\)

\(\Leftrightarrow6x^2-12x-\left(6x^2-10x\right)=44\Leftrightarrow-2x=44\Leftrightarrow x=-22\)

c, \(2\left(5x-8-3\right)\left(4x-5\right)=4\left(3x-4\right)\)

\(\Leftrightarrow\left(10x-22\right)\left(4x-5\right)=4\left(3x-4\right)\Leftrightarrow40x^2-50x-88x+110=0\)

\(\Leftrightarrow40x^2-138x+110=0\)( vô nghiệm ) 

a,\(\frac{1}{3}x=-2-\frac{2}{3}=\frac{-8}{3}\)

        \(x=\frac{-8}{3}:\frac{1}{3}=\frac{-8}{3}.\frac{3}{1}=-8\)

\(b,\left[x+\frac{1}{1}\right]^2+\frac{5}{6}=\frac{7}{8}\)

\(\Rightarrow\left[x+1\right]^2=\frac{7}{8}-\frac{5}{6}\)

\(\Rightarrow\left[x+1\right]^2=\frac{7\cdot3}{24}-\frac{5\cdot4}{24}\)

\(\Rightarrow\left[x+1\right]^2=\frac{21}{24}-\frac{20}{24}\)

\(\Rightarrow\left[x+1\right]^2=\frac{1}{24}\)

\(\Rightarrow x\in\left\{\varnothing\right\}\)

\(c,\left[3x+\frac{3}{5}\right]\left[\left|x\right|-\frac{1}{4}\right]=0\)

\(\Rightarrow\hept{\begin{cases}3x+\frac{3}{5}=0\\\left|x\right|-\frac{1}{4}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3x=0-\frac{3}{5}\\\left|x\right|=0+\frac{1}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3x=\frac{-3}{5}\\\left|x\right|=0+\frac{1}{4}\\\left|x\right|=0+\left[\frac{-1}{4}\right]\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{5}\\x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{5};\pm\frac{1}{4}\right\}\)

a: đặt \(x^2-2\left(x^2-8\right)=0\)

\(\Leftrightarrow16-x^2=0\)

=>x=4 hoặc x=-4

b: Đặt \(3x-5-4\left(2x+3\right)=0\)

=>3x-5-8x-12=0

=>-5x-17=0

=>-5x=17

hay x=-17/5

c: Đặt \(3y^2-5y=0\)

=>y(3y-5)=0

=>y=0 hoặc y=5/3

d: Đặt \(2x^2-3\left(x^2+4\right)=0\)

\(\Leftrightarrow-x^2-12=0\)

hay \(x\in\varnothing\)

a: \(\Leftrightarrow\left[\left(3x+14\right):4-3\right]:2=1\)

=>(3x+14):4-3=2

=>(3x+14):4=5

=>3x+14=20

=>3x=6

hay x=2

b: \(\Leftrightarrow\left[\left(x:4+17\right):10+3\cdot16\right]:10=5\)

\(\Leftrightarrow\left(x:4+17\right):10=50-48=2\)

=>x:4+17=20

=>x:4=3

hay x=12

c: \(\Leftrightarrow2\cdot15^2+\left[2\cdot125-\left(2x+4\right)\cdot5\right]:19=453\)

\(\Leftrightarrow250-\left(2x+4\right)\cdot5=\left(453-450\right)\cdot19=57\)

=>5(2x+4)=197

=>2x+4=197/5

=>2x=177/5

hay x=177/10

d: \(\Leftrightarrow\left(19x+50\right):14=5^2-4^2=9\)

=>19x+50=126

=>19x=76

hay x=4

e: \(\Leftrightarrow2\cdot3^x=10\cdot3^{12}+8\cdot3^{12}=18\cdot3^{12}\)

\(\Leftrightarrow3^x=3^2\cdot3^{12}=3^{14}\)

hay x=14

f: \(\Leftrightarrow3\left(x+2\right):7=30\)

=>3(x+2)=210

=>x+2=70

hay x=68

g: \(2480-1570+200-x+5=1010\)

=>1115-x=1010

hay x=105