Tìm x
a) \(2^{3x+2}=4^{x+5}\)
b)\(3^{-1}.3^x+9.3^x=28\)
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(x-1)(x+2)(x+3)(x+6)
= [(x-1)(x+6)].[(x+2)(x+3)]
=(x^2+5x-6)(x^2+5x+6)
=(x^2+5x)^2 -6^2 = (x^2+5x)^2 -36
vì (x^2+5x)^2 > hoặc bằng 0 => (x-1)(x+2)(x+3)(x+6) > hoặc bằng -36.
Dấu bằng xảy ra khi (x^2+5x)^2=0 <=> x=0 hoặc x= -5
c) \(x^2-9=2\cdot\left(x+3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
d) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)
\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)
a) \(x^2-9=2\left(x+3\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)
\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
c) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
câu a giống Võ Đoan Nhi
câu b:
( x2 + 2x -11 ) : ( x + 2)
=> x2 + 2x -11 : ( x + 2)
=> x(x+2) -11 : ( x + 2)
Vì x( x + 2) : ( x + 2) nên -11 : ( x + 2)
=> x + 2 thuộc ước của -11
ta lập bảng..............
\(3x+4⋮x-3\)
\(\Leftrightarrow3\left(x-3\right)+10\)\(⋮x-3\)
-Mà: \(3\left(x-3\right)⋮x-3\Rightarrow10⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(10\right)\Leftrightarrow x-3\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
-Lập bảng:.....
a, 3 - 2 | 5x - 4 | = -11
2|5x - 4| = 14
|5x - 4| = 7
Th1: 5x -4 =7
5x = 11
x= 11/5
Th2:
5x -4 =-7
5x = -3
x= -3/5
a) => 2/5x-4/=14
=> /5x-4/=7
=> 5x-4=7 hoac 5x-4=-7
x=11/5 x=-3/5
a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)
b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)
c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)
\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)
Vì -50<-30<4<5
nên A<D<B<C
a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)
b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)
c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)
\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)
Vì -50<-30<4<5
nên A<D<B<C
a) 23x+2 = 4x+5 = (22)x+5 = 22x+10
=> 3x + 2 = 2x + 10
=> 3x - 2x = 10 - 2
x = 8
b) 3-1.3x + 9.3x = 28
3x. ( 3-1 + 9) = 28
3x. 28/3 = 28
3x = 3 = 31
=> x = 1
\(2^{3x+2}=4^{x+5}\)
\(\Rightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Rightarrow3x+2=2\left(x+5\right)\)
\(\Rightarrow3x+2=2x+10\)
\(\Rightarrow3x-2x=10-2\Rightarrow x=8\)
\(3^{-1}.3^x+9.3^x=28\)
\(\Rightarrow\frac{1}{3}.3^x+9.3^x=28\)
\(\Rightarrow3^x.\left(9+\frac{1}{3}\right)=28\)
\(\Rightarrow3^x.\frac{28}{3}=28\)
\(\Rightarrow3^x=3\Rightarrow x=1\)
Chúc bạn học tốt.