Tìm x,biết
3^x+3^x+1+3^x+2+3^x+3=116
Help me ;-)♥
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Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
\(\frac{x}{3}+\frac{x^2}{2}=0\)
\(\Leftrightarrow\frac{2x+3x^2}{6}=0\Leftrightarrow3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{2}{3}\end{cases}}\)
\(\left(x^2+3\right)\left(x+1\right)+x=-1\)
\(\Leftrightarrow\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(x+1\right)=0\)
Mà \(x^2+4>0\)nên \(x+1=0\Leftrightarrow x=-1\)
a) x+(x+1)+(x+2)+(x+3)+...+(x+30)=1240
( x + x + x + x + ... + x ) + ( 1 + 2 + 3 +... + 30 ) = 1240
31x + 465 = 1240
31x = 1240 - 465
31x = 775
x = 775 : 31
x = 25
a) x+(x+1)+(x+2)+(x+3)+...+(x+30)=1240
( x + x + x + x + ... + x ) + ( 1 + 2 + 3 +... + 30 ) = 1240
31x + 465 = 1240
31x = 1240 - 465
31x = 775
x = 775 : 31
x = 25
5x(x-3)^2-5(x-1)^3+15(x+2)(x-2)=5
5x(x-3)^2-5(x-1)^3+15(x^2-2^2)=5
5x(x^2-6x+9)-5(x^3-3x^2+3x-1)+15x^2-60=5
5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5
30x-55=5
30x=60
x=2
a: \(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}+\dfrac{4x^2}{x^2-9}\right):\dfrac{2x+1-x-3}{x+3}\)
\(=\dfrac{-x^2-6x-9+x^2-6x+9+4x^2}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x-2}\)
\(=\dfrac{4x^2-12x}{x-3}\cdot\dfrac{1}{x-2}=\dfrac{4x}{x-2}\)
b: \(2x^2-5x+2=0\)
=>(x-2)(2x-1)=0
=>x=1/2
Thay x=1/2 vào P, ta được:
\(P=\left(4\cdot\dfrac{1}{2}\right):\left(\dfrac{1}{2}-2\right)=2:\dfrac{-3}{2}=\dfrac{-4}{3}\)
\(3^x+3^{x+1}+3^{x+2}+3^{x+3}=116\)
\(\Leftrightarrow3^x+3^x.3+3^x.3^2+3^x.3^3=116\)
\(\Leftrightarrow3^x\left(1+3+3^2+3^3\right)=116\)
\(\Leftrightarrow3^x.\left(4+9+27\right)=116\)
\(\Leftrightarrow3^x.40=116\)
\(\Leftrightarrow3^x=2,9\)
Tự làm nốt nhế:)))) _____ Nghi là sai đề
@@ Học tốt
\(3^x+3^{x+1}+3^{x+2}+3^{x+3}=116\)
\(3^x+3^x.3^1+3^x.3^2+3^x.3^3=116\)
\(3^x\left(1+3+3^2+3^3\right)=116\)
\(3^x.40=116\)
\(3^x=116:40\)
\(3^x=2,9\)
thấy đề sai sai