64^4 - x^3 - 7x^2 + x +1=0
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24 tháng 8 2021
Ta có: \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+x\left(7x-6\right)=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+7x^2-6x=0\)
\(\Leftrightarrow x^2+7x-8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)
23 tháng 12 2021
\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{10}{7}\end{matrix}\right.\)
DN
23 tháng 12 2021
\(\left(x-20\right).\left(7x+10\right)=0\)
\(=>\left[{}\begin{matrix}x-20=0\\7x+10=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=20\\x=-\dfrac{10}{7}\end{matrix}\right.\)
10 tháng 11 2023
a: \(\lim\limits_{x\rightarrow-2}\dfrac{4-x^2}{2x^2+7x+6}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(2-x\right)\left(2+x\right)}{2x^2+4x+3x+6}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(2-x\right)\left(x+2\right)}{\left(x+2\right)\left(2x+3\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{2-x}{2x+3}=\dfrac{2-\left(-2\right)}{2\cdot\left(-2\right)+3}=\dfrac{4}{-4+3}=-4\)
b: \(\lim\limits_{x\rightarrow4}\dfrac{2x^2-13x+20}{x^3+64}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2x^2-8x-5x+20}{\left(x+4\right)\left(x^2-4x+16\right)}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{\left(x-4\right)\left(2x-5\right)}{x^3+64}\)
\(=\dfrac{\left(4-4\right)\left(2\cdot4-5\right)}{4^3+64}=0\)
c: \(\lim\limits_{x\rightarrow-1}\dfrac{2x^2+8x+6}{-2x^2+7x+9}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{2x^2+2x+6x+6}{-2x^2-2x+9x+9}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x+6\right)}{-2x\left(x+1\right)+9\left(x+1\right)}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x+6\right)}{\left(x+1\right)\left(-2x+9\right)}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{2x+6}{-2x+9}=\dfrac{2\cdot\left(-1\right)+6}{-2\cdot\left(-1\right)+9}\)
\(=\dfrac{4}{11}\)
N
1
17 tháng 12 2017
1)7x(x-5)-x(x-5)=(x-5)(7x-x)=6x(x-5)
2)x4+3x3+x+3=x3(x+3)+(x+3)=(x+3)(x3+1)=(x+3)(x+1)(x2-x+1)
3)x4+64=[(x2)2+2.x2.8+64]-16x2=(x2+8)2-(4x)2=(x2+4x+8)(x2-4x+8)
VD
29 tháng 5 2022
\(\dfrac{1}{3}x+\dfrac{2}{3}\left(x-1\right)=0\\ \dfrac{1}{3}x+\dfrac{2}{3}x-\dfrac{2}{3}=0\\ x=\dfrac{2}{3}\)
25 tháng 10 2023
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
28 tháng 7 2023
2:
a: =>(x-9)(x-1)=0
=>x=9 hoặc x=1
b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0
=>(x+4)(x^2-4x+16+x-16)=0
=>(x+4)(x^2-3x)=0
=>x(x-3)(x+4)=0
=>x=0;x=3;x=-4
YN
28 tháng 7 2023
bài 2 :
a: =>(x-9)(x-1)=0
=>x=9 hoặc x=1
b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0
=>(x+4)(x^2-4x+16+x-16)=0
=>(x+4)(x^2-3x)=0
=>x(x-3)(x+4)=0
=>x=0;x=3;x=-4
