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3x+3x+1+3x+2=117
=> 3x.(1+3+32)=117
=> 3x.(1+3+9)=117
=> 3x.13=117
=> 3x=117:13
=> 3x=9
=> 3x=32
=> x=2
3x+3x+1+3x+2=117
=> 3x.(1+3+32)=117
=> 3x.(1+3+9)=117
=> 3x.13=117
=> 3x=117:13
=> 3x=9
=> 3x=32
=> x= 2
tick nha
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\(3^x+3^{x+1}+3^{x+2}=117\)
\(\Leftrightarrow3^x+3^x.3^1+3^x.3^2=117\)
\(\Leftrightarrow3^x\left(1+3+3^2\right)=117\)
\(\Leftrightarrow3^x.13=117\)
\(\Leftrightarrow3^x=9=3^2\)
\(\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\Rightarrow3^x\left(3^2+3+1\right)=117\)
\(\Rightarrow3^x.13=117\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
=>x=2
b)
\(3^{2x+1}=3^{-4}\)
=> 2x+1= - 4
=>\(x=-\frac{5}{2}\)
c)
\(\left(x+2\right)^4=16\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x+2\right)^4=2^4\\\left(x+2\right)^4=\left(-2\right)^4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=2\\x+2=-2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\-4\end{array}\right.\)
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\(3^x+3^{x-1}+3^{x-2}=117\)
\(\Leftrightarrow3^x+\frac{3^x}{3}+\frac{3^x}{3^2}=117\)
\(\Leftrightarrow3^x.\left(1+\frac{1}{3}+\frac{1}{9}\right)=117\)
\(\Leftrightarrow3^x.\frac{13}{9}=117\)
\(\Leftrightarrow3^x=81\)
\(\Leftrightarrow3^x=3^4\)
\(\Leftrightarrow x=4\)
~Học tốt~
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Nhân phân phối là ra thôi
a)
\(VT=\left(x-1\right)\left(x+1\right)=x.x+x.1-1.x+\left(-1\right).1\)
\(=\left(x^2-1\right)+\left(x-x\right)=x^2-1+0=x^2-1=VP\Rightarrow dccm\)
c) thay vì c/m A=B ta chứng Minh B=A
\(VP=\left(x+1\right)\left(x^2-x+1\right)=\left(x^3-x^2+x\right)+\left(x^2-x+1\right)\)
\(=\left(x^3+1\right)+\left(-x^2+x^2\right)+\left(x-x\right)=x^3+1+0+0=x^3+1=VT\Rightarrow VT=VP\Rightarrow dpcm\)\(=x^3+1+0+0=x^3+1=VT\Rightarrow VT=VP\Rightarrow dpcm\)
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\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{6}\rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{4}=\dfrac{z^2}{36}\rightarrow\dfrac{5x^2}{45}=\dfrac{y^2}{4}=\dfrac{z^2}{36}\)
-Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{5x^2}{45}=\dfrac{y^2}{4}=\dfrac{z^2}{36}=\dfrac{5x^2+y^2-z^2}{45+4-36}=\dfrac{117}{13}=9\)
-Suy ra: x2=9.9=81\(\rightarrow\)x=\(\pm9\)
y2=9.4=36\(\rightarrow\)y=\(\pm6\)
z2=9.36=324\(\rightarrow\)z=\(\pm18\)
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{6}\) và \(5x^2+y^2-z^2=117\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{6}=k\)
=> \(\left\{{}\begin{matrix}x=3k\\y=2k\\z=6k\end{matrix}\right.\)
Ta có:
5x2 + y2 - z2 = 117
=> 5(3k)2 + (2k)2 - (6k)2 = 117
=> 5.9k2 + 4k2 - 36k2 = 117
=> 45k2 + 4k2 - 36k2 = 117
=> (45 + 4 - 36).k2 = 117
=> 13k2 = 117
=> k2 = 117 : 13
=> k2 = 9 => \(\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)
*Với k = 3 ta có:
x = 3.3 = 9 ; y = 3.2 = 6 ; z = 3.6 = 18
*Với k = -3 ta có:
x = -3.3 = -9 ; y = -3.2 = -6 ; z = -3.6 = -18
\(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x\left(1+3+3^2\right)=117\)
\(3^x=\dfrac{117}{13}=9=3^2\)
\(\Rightarrow x=2\)
Cám ơn bạn nha