a) 16x : 4x = 16
b) \(2^{-1}.2^x+4.2^x=72\)
c) \(\left(2^x+1\right)^3=-64\)
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10 tháng 11 2023
a: \(P=-3x^3+5x\)
\(=x\cdot\left(-3x^2\right)+x\cdot5\)
\(=x\left(-3x^2+5\right)\)
b: \(Q=\left(2x-1\right)+\left(x-2\right)\left(2x-1\right)\)
\(=\left(2x-1\right)\left(1+x-2\right)\)
\(=\left(2x-1\right)\left(x-1\right)\)
c: \(R=4-16x^2\)
\(=4\cdot1-4\cdot4x^2\)
\(=4\left(1-4x^2\right)\)
\(=4\left(1-2x\right)\left(1+2x\right)\)
d: \(S=36-4x^2\)
\(=4\cdot9-4\cdot x^2\)
\(=4\left(9-x^2\right)\)
\(=4\left(3-x\right)\left(3+x\right)\)
e: \(T=8x^3-1\)
\(=\left(2x\right)^3-1^3\)
\(=\left(2x-1\right)\left(4x^2+2x+1\right)\)
f: \(Q=8-x^3\)
\(=2^3-x^3\)
\(=\left(2-x\right)\left(4+2x+x^2\right)\)
g: \(N=64-x^3\)
\(=4^3-x^3\)
\(=\left(4-x\right)\left(16+4x+x^2\right)\)
15 tháng 9 2021
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
15 tháng 9 2021
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
1 tháng 9 2020
a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
HH
17 tháng 6 2018
a/ \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
<=> \(48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
<=> \(83x-2=81\)
<=> \(83x=83\)
<=> \(x=1\)
b/ \(\left(2x-3\right)\left(2x+3\right)-\left(4x+1\right)x=1\)
<=> \(4x^2-9-4x^2-x=1\)
<=> \(-\left(9+x\right)=1\)
<=> \(9+x=-1\)
<=> \(x=-10\)
c/ \(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
<=> \(3x^2-\left(3x^2-x+6x-2\right)=-7\)
<=> \(3x^2-3x^2+x-6x+2=-7\)
<=> \(-5x+2=-7\)
<=> \(-5x=-9\)
<=> \(x=\frac{9}{5}\)
KN
31 tháng 3 2018
Câu a và câu c bn kia làm rồi nên mk làm câu b thôi nhé....
b) y2 + 4x + 2y - 2x+1 + 2 = 0
\(\Leftrightarrow\) (y2 + 2y + 1) + 4x - 2x.2 + 1 = 0
\(\Leftrightarrow\) (y + 1)2 + [(2x)2 - 2.2x.1 + 1] = 0
\(\Leftrightarrow\) (y + 1)2 + (2x - 1)2 = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}y+1=0\\2^x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\2^x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\x=0\end{matrix}\right.\)
Vậy...................
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2019
a/
\(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2+2\left(x^2+2x\right)+1=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Rightarrow x=1\)
b/
\(y^2+2y+1+\left(2^x\right)^2-2.2^x+1=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\2^x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-1\\x=0\end{matrix}\right.\)
c/
ĐKXĐ: \(x\ne\left\{-2;-4;-6;-8\right\}\)
\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\frac{2}{x+2}+x+8+\frac{8}{x+8}=x+4+\frac{4}{x+4}+x+6+\frac{6}{x+6}\)
\(\Leftrightarrow\frac{1}{x+2}+\frac{4}{x+8}=\frac{2}{x+4}+\frac{3}{x+6}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{2}{x+4}+\frac{4}{x+8}-\frac{3}{x+6}=0\)
\(\Leftrightarrow\frac{-x}{\left(x+2\right)\left(x+4\right)}+\frac{x}{\left(x+8\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{\left(x+2\right)\left(x+4\right)}=\frac{1}{\left(x+6\right)\left(x+8\right)}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x+2\right)\left(x+4\right)=\left(x+6\right)\left(x+8\right)\)
\(\Leftrightarrow8x=-40\Rightarrow x=-5\)
PA
31 tháng 1 2017
a) (−11)2 − 15(x − 2) = 134 − 16x
121 - 15x + 30 = 134 - 16x
16x - 15x = 134 - 121 - 30
x = -17
b) (4x + 1)(x2 − 16)=0
(4x + 1)(x - 4)(x + 4) = 0
\(\left[\begin{matrix}4x+1=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\left[\begin{matrix}4x=-1\\x=4\\x=-4\end{matrix}\right.\)
\(\left[\begin{matrix}x=-\frac{1}{4}\\x=4\\x=-4\end{matrix}\right.\)
c) − 2(x − 3) + (− 2)2 = 4 − 3x
3x + 4 - 2x + 6 = 4
x = 4 - 4 - 6
x = - 6
\(a,16^x:4^x=16\)
\(\left(4^x\right)^2:4^x=4^2\)
\(\Rightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,2^{-1}.2^x+4.2^x=72\)
\(\Rightarrow2^{x-1}+2^{x+2}=72\)
\(\Rightarrow2^{x-1}\left(1+2^3\right)=72\)
\(\Rightarrow2^{x-1}=72:9=8=2^3\)
\(\Rightarrow x=4\)
\(c,\left(2^x+1\right)^3=-64\)
\(\Rightarrow2^x+1=-4\)
\(\Rightarrow2^x=-5\)
a) 16^x : 4^x = 16
<=> ( 16 : 4 )^x = 16
<=> 4^x = 16
<=> 4^x = 4^2
=> x = 2
Vậy x =2