Tìm x biết :
a) x2\(-\)4x+4=25
b) \(\dfrac{x-17}{1990}\)+\(\dfrac{x-21}{1986}\)+\(\dfrac{x+1}{1004}\)=4
c) 4x\(-\)12.2x+32=0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
28 tháng 2 2018
a, <=> (x-2)2=25
<=>x-2=5 hoặc x-2=-5
<=>x=7 hoặc x=-3
c,<=>(x2)2-16=0
<=>(x2)2=16
<=>x2=4
<=>x=2 hoặc x=-2
HH
3
25 tháng 4 2017
\(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)
\(\Rightarrow\left(\dfrac{x-17}{1990}-1\right)+\left(\dfrac{x-21}{1986}-1\right)+\left(\dfrac{x+1}{1004}-2\right)=0\)
\(\Rightarrow\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)
\(\Rightarrow\left(x-2007\right)\left(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\right)=0\)
\(\Rightarrow x-2007=0\Rightarrow x=2007\)
Vậy x = 2007
TH
0
MA
0
5 tháng 1 2021
a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)
b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)
c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)
NH
11 tháng 3 2023
bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu
11 tháng 3 2023
a: =>\(4x-5=2x-2+x=3x-2\)
=>x=3
b: \(\Leftrightarrow7x-35=3x+6\)
=>4x=41
=>x=41/4
c: =>(2x+5)(x+5)-2x^2=0
=>2x^2+10x+5x+25-2x^2=0
=>15x=-25
=>x=-5/3
e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)
=>11(x^2-3x-4)=x(11x-34)
=>11x^2-33x-44=11x^2-34x
=>x=44
a)
\(x^2-4x+4=25\)
\(\Leftrightarrow x^2-4x-21=0\)
\(\Leftrightarrow x^2+3x-7x-21=0\)
\(\Leftrightarrow x\left(x+3\right)-7\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b)
\(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)
\(\Leftrightarrow\dfrac{x-17}{1990}-1+\dfrac{x-21}{1986}-1+\dfrac{x+1}{1004}-2=4-1-1-2\)
\(\Leftrightarrow\dfrac{x-17-1990}{1990}+\dfrac{x-21-1986}{1986}+\dfrac{x+1-2008}{1004}=0\)
\(\Leftrightarrow\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\right)=0\)
\(\Leftrightarrow x-2007=0\) ( Vì: \(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\ne0\))
\(\Leftrightarrow x=2007\)
c.
\(4^x-12.2^x+32=0\)
\(\Leftrightarrow\left(2^x\right)^2-12.2^x+36-4=0\)
\(\Leftrightarrow2^x-2.2^x.6+6^2-2^2=0\)
\(\Leftrightarrow\left(2^x-6\right)^2-2^2=0\)
\(\Leftrightarrow\left(2^x-6-2\right)\left(2^x-6+2\right)=0\)
\(\Leftrightarrow\left(2^x-8\right)\left(2^x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x-8=0\\2^x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x=8\\2^x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)