a) \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)

⇔ | 3 - 2x | + 3/4 = 11/4

⇔ | 3 - 2x | = 8/4 = 2

⇔ \(\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}}\text{⇔}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)

b) 2^x+2 - 2^x = 96

⇔ 2^x( 2² - 1 ) = 96

⇔ 2^x.3 = 96

⇔ 2^x = 32

⇔ 2^x = 2⁵

⇔ x = 5

c) ( 2x + 5 )³ = -27

⇔ ( 2x + 5 )³ = (-3)³

⇔ 2x + 5 = -3

⇔ 2x = -8

⇔ x = -4

a. \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)

\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\left|-\frac{11}{4}\right|\)

\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\frac{11}{4}\)

\(\Rightarrow\left|3-2x\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)

b. 2^x+2 - 2^x = 96

<=> 2^x . 2² - 2^x = 96

<=> 2^x ( 2² - 1 ) = 96

<=> 2^x . 3 = 96

<=> 2^x = 32 = 2⁵

<=> x = 5

c. ( 2x + 5 )³ = - 27

<=> ( 2x + 5 )³ = ( - 3 )³

<=> 2x + 5 = - 3

<=> 2x = - 8

<=> x = - 4

\(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|=>\left|3-2x\right|+\frac{3}{4}=\left|-\frac{11}{4}\right|\)

\(\left|3-2x\right|+\frac{3}{4}=\frac{11}{4}=>\left|3-2x\right|=\frac{11}{4}-\frac{3}{4}\)

\(\left|3-2x\right|=2=>\orbr{\begin{cases}\left|3-2x\right|=2\\\left|3-2x\right|=-2\end{cases}}=>\orbr{\begin{cases}2x=2+3\\2x=-2+3\end{cases}}\)

\(\orbr{\begin{cases}2x=5\\2x=1\end{cases}}=>\orbr{\begin{cases}x=5:2\\x=1:2\end{cases}}=>\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{1}{2}\end{cases}}\)

vậy \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{1}{2}\end{cases}}\)

\(2^{x+2}-2^x=96=>2^x.2^2-2^x=96=>2^x.\left(2^2-1\right)=96\)

\(2^x.\left(4-1\right)=96=>2^x.3=96=>2^x=96:3=>2^x=32\)

\(2^x=2^5=>x=5\)

vậy \(x=5\)

\(\left(2x+5\right)^3=-27=>\left(2x+3\right)^3=\left(-3\right)^3=>2x+3=-3\)

\(2x=-3-3=>2x=-6=>x=-6:3=>x=-2\)

vậy \(x=-2\)

Xét cấp số cộng 1, 6, 11, ..., 96.

Ta có: 96 = 1 + 5(n − 1) ⇒ n = 20

Suy ra

Và 2x.20 + 970 = 1010

Từ đó x = 1

a: =>12x-64=32

=>12x=96

=>x=8

b: =>x-1=5

=>x=6

c: =>2^x*3=96

=>2^x=32

=>x=5

a) Ta có : |2x - 5| = x + 1

\(\Leftrightarrow\orbr{\begin{cases}2x-5=-x-1\\2x-5=x+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+5\\2x-x=1+5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=4\\x=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=6\end{cases}}\)

mik ko pc

\(a,\Rightarrow\left(35x+3\right)\cdot19=152\\ \Rightarrow35x+3=8\\ \Rightarrow x=\dfrac{1}{7}\\ b,\Rightarrow3\left(x+7\right)=42\\ \Rightarrow x+7=14\Rightarrow x=7\\ c,\Rightarrow3\left(x+1\right)=48\\ \Rightarrow x+1=16\Rightarrow x=15\\ d,\Rightarrow120-5x+100\cdot2:5=4\cdot15\\ \Rightarrow120-5x+40=60\\ \Rightarrow5x=100\Rightarrow x=20\\ e,\Rightarrow4x-10=30\\ \Rightarrow4x=40\\ \Rightarrow x=10\\ g,\Rightarrow10x+10=70\\ \Rightarrow10x=60\\ \Rightarrow x=6\)

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)

\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)