\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\Rightarrow x-\dfrac{1}{2}=\dfrac{2}{5}\)

\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{2}\Rightarrow x=\dfrac{9}{10}\)

\(2^x+2^{x+3}=144\Rightarrow2^x+2^x.2^3=144\Rightarrow2^x\left(1+2^3\right)=144\Rightarrow9.2^x=144\Rightarrow2^x=144:9=16\Rightarrow2^x=2^4\Rightarrow x=4\)

thank you!

Thật ra câu này mk làm rồi nhưng chưa chắc chắn cho lắm!

2^x + 2^{x + 3} = 144

2^x . 1 + 2^x . 2³ = 144

2^x . (1 + 8) = 144

2^x . 9 = 144

2^x = 16 = 2⁴

=> x = 4

\(2^x+2^{x+3}=144\)

\(\Leftrightarrow2^x+2^x.2^3=144\)

\(\Leftrightarrow2^x+2^x.8=144\)

\(\Leftrightarrow2^x.\left(1+8\right)=144\)

\(\Leftrightarrow2^x.9=144\)

\(\Leftrightarrow2^x=16\)\(\Leftrightarrow2^x=2^4\)

\(\Leftrightarrow x=4\)

Vậy \(x=4\)

Các bạn ơi! Dấu chấm là dấu nhân nha!

Ta có: \(144=2^4.3^2.5^0\)

Suy ra: \(2^{x-2}.3^{y-3}.5^{z-1}=2^4.3^2.5^0\)

Suy ra: \(2^{x-2}=2^4;3^{y-3}=3^2;5^{z-1}=5^0\)

Suy ra: \(x-2=4;y-3=2\) và \(z-1=0\)

Hay \(x=6;y=5\) và \(z=1\)

a) \(81^{-2x}.27^x=9^5\)

\(\Rightarrow\left(3^4\right)^{-2x}.\left(3^3\right)^x=3^{10}\)

\(\Rightarrow3^{-8x}.3^{3x}=3^{10}\)

\(\Rightarrow3^{-5}=3^{10}\)

\(\Rightarrow-5x=10\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

b) \(2^x+2^{x+3}=144\)

\(\Rightarrow\left(1+2^3\right).2^x=144\)

\(\Rightarrow\left(1+8\right).2^x=144\)

\(\Rightarrow9.2^x=144\)

\(\Rightarrow2^x=16\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

c) \(2^{x-1}+5.2^{x-2}=7.32\)

\(\Rightarrow\left(2+5\right).2^{x-2}=244\)

\(\Rightarrow7.2^{x-2}=244\)

\(\Rightarrow2^{x-2}=32\)

\(\Rightarrow x-2=5\)

\(\Rightarrow x=7\)

Vậy \(x=7\)

Giải

a, Ta có 2^x + 2^x+5 = 144

=> 2^x.1 + 2^x.2^5 = 144

=> 2^x.(1+2^5)=144

=> 2^x.33=144

=> 2^x=144/33=48/11

Vì 2^x luôn dương mà 48/11 là một phân số 

=> Vô lý

Vậy không tìm được giá trị x thỏa mãn

b, Giải

Ta có |x+1|+|x+3|+|x+5|=7x

=> x+1+x+3+x+5=7x

=> 3x+9=7x

=> 9=7x-3x

=>9=4x

=> 9/4=x

Vậy x=9/4

a)\(\frac{10^3+2.5^3+5^3}{55}\)=\(\frac{5^3.2^3+2.5^3+5^3}{5.11}\)=\(\frac{5^3.\left(2^3+2+1\right)}{5.11}\)=\(5^2\)=\(25\)

b) \(2^x+2^{x+3}=144\)

\(\Rightarrow2^x+2^x.2^3=144\)

\(\Rightarrow2^x.\left(1+2^3\right)=144\)

\(\Rightarrow2^x=16=2^4\)

\(\Rightarrow x=4\)

c) \(2\left(x-5\right)+3\left(x-7\right)=10\)

\(\Rightarrow2x-10+3x-21=10\)

\(\Rightarrow5x-31=10\)

\(\Rightarrow5x=41\)

\(\Rightarrow x=\frac{41}{5}=8,2\)

a) \(2^x+2^{x+5}=144\)

\(\Rightarrow2^x+2^x\cdot2^5=144\)

\(\Rightarrow2^x+2^x\cdot32=144\)

\(\Rightarrow2^x\left(1+32\right)=144\)

\(\Rightarrow2^x\cdot33=144\)

\(\Rightarrow2^x=144:33\)

\(\Rightarrow2^x=\frac{48}{11}\)

\(\Rightarrow x\in\varnothing\)

Vậy không tìm được x thỏa mãn đề bài

b) \(|x+1|+|x+3|+|x+5|=7x\)

Ta có: \(\hept{\begin{cases}|x+1|\ge0\forall x\\|x+3|\ge0\forall x\\|x+5|\ge0\forall x\end{cases}\Rightarrow|x+1|+|x+3|+|x+5|\ge0\forall x\Rightarrow7x\ge0\forall x}\)

\(\Rightarrow|x+1|+|x+3|+|x+5|=x+1+x+3+x+5=7x\)

\(\Rightarrow\left(x+x+x\right)+\left(1+3+5\right)=7x\)

\(\Rightarrow3x+9=7x\)

\(\Rightarrow7x-3x=9\)

\(\Rightarrow4x=9\)

\(\Rightarrow x=\frac{4}{9}\)

Vậy x=\(\frac{4}{9}\)

\(\left|x+1\right|+\left|x+3\right|+\left|x+5\right|=7x^{\left(1\right)}\)

Ta có \(\left|x+1\right|\ge0;\left|x+3\right|\ge0;\left|x+5\right|\ge0\)

\(\Rightarrow7x\ge0\Rightarrow x\ge0\)

Từ (1)\(\Rightarrow\left|x+1\right|+\left|x+3\right|+\left|x+5\right|=7x\)

\(3x+9=7x\)

\(3x-7x=-9\)

\(-4x=-9\)

\(x=\frac{9}{4}\)