( 2x - 1) mũ 3 = 125
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\(\left(x+1\right)^3=125\)
\(\left(x+1\right)^3=\sqrt[3]{125}\)
\(\left(x+1\right)=5\)
\(x=5-1\)
\(x=4\)
\(2^x+2^{x+3}=576\)
\(2^x\left(1+2^3\right)=576\)
\(2^x=576:9\)
\(2^x=64\)
\(2^x=2^6\)
\(\Rightarrow x=6\)
Bài 4:
a: \(=7xy\left(2-3-4\right)=-35xy\)
b: \(=\left(x-5\right)\left(x+y\right)\)
c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)
d: \(=\left(x+y\right)^3-\left(x+y\right)\)
=(x+y)(x+y+1)(x+y-1)
e: =x^2+8x-x-8
=(x+8)(x-1)
f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)
g: =-5x^2+15x+x-3
=(x-3)(-5x+1)
h: =x^2-3xy+xy-3y^2
=x(x-3y)+y(x-3y)
=(x-3y)*(x+y)
Bài 4:
a: \(=7xy\left(2-3-4\right)=-35xy\)
b: \(=\left(x-5\right)\left(x+y\right)\)
c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)
d: \(=\left(x+y\right)^3-\left(x+y\right)\)
=(x+y)(x+y+1)(x+y-1)
e: =x^2+8x-x-8
=(x+8)(x-1)
f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)
g: =-5x^2+15x+x-3
=(x-3)(-5x+1)
h: =x^2-3xy+xy-3y^2
=x(x-3y)+y(x-3y)
=(x-3y)*(x+y)
Bài 1:
a: Ta có: |3x-2|+|2y+1|=0
=>3x-2=0 và 2y+1=0
=>x=2/3 và y=-1/2
Bài 2:
a: ta có: \(\left(2x-5\right)^{x-3}=\left(2x-5\right)^2\)
\(\Leftrightarrow\left(2x-5\right)^{x-3}-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-5\right)^2\left[\left(2x-5\right)^{x-5}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\x-5=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{5}{2};5\right\}\)
b: Ta có; \(x^{2x-1}=x^3\)
\(\Leftrightarrow x^3\left(x^{2x-4}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-4=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;2\right\}\)
Câu 2:
Ta có: \(9-7x^2=30\)
\(\Leftrightarrow7x^2=9-30=-21\)(Vô lý)
\(\left(2x-1\right)^3=125\\ \Rightarrow2x-1=5\\ \Rightarrow2x=6\\ \Rightarrow x=3\)
(2x−1)3 = 125
⇒ 2x − 1 = 5
⇒ 2x = 6
⇒ x = 3