(2|x|+1).(8x^(3)+1)=0
(x^2-4)^2=|x+2|=0
giúp mik vs,mik cần gấp;-;;;
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a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
giải giúp mik với a) 2^x+1 =64
b) 570-x: 3 và 17<x<20
c) (4x-9)-(x+111)=0
giúp mik với nha mik cần gấp
\(a,2^{x+1}=64\\ \Rightarrow a,2^{x+1}=2^6\\ \Rightarrow x+1=6\\ \Rightarrow x=5\)
\(b,x=18\)
\(c,\left(4x-9\right)-\left(x+111\right)=0\\ \Rightarrow4x-9-x-111=0\\ \Rightarrow3x-120=0\\ \Rightarrow3x=120\\ \Rightarrow x=40\)
\(x\left(x-\frac{1}{3}\right)< 0\)
Để \(x\left(x-\frac{1}{3}\right)< 0\)thì x và \(x-\frac{1}{3}\)trái dấu nhau
Thấy \(x>x-\frac{1}{3}\)\(\Rightarrow\hept{\begin{cases}x>0\\x-\frac{1}{3}< 0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x< \frac{1}{3}\end{cases}\Leftrightarrow}0< x< \frac{1}{3}}\)
a, | x - 3/4 | = 1/2
=>\(\orbr{\begin{cases}x-\frac{3}{4}=\frac{1}{2}\\x-\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{2}{4}+\frac{3}{4}\\x=-\frac{2}{4}+\frac{3}{4}\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)
Vậy....
a) \(|x-\frac{3}{4}|=\frac{1}{2}\)
\(< =>\orbr{\begin{cases}x-\frac{3}{4}=\frac{1}{2}\\x-\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)
Vay : x = 5/4 hoặc x = 1/4
b)\(saide\)
tìm x biết :
a,(2x-3)^2 =(x+ 5)^2
b,x^2(x-1) -4x^2 +8x -4 =0
c, (x-4)^2 -36 =0
giúp mik nha mik đang gấp
a, (2x-3)^2=(x+5)^2
2x-3=x+5
2x-3-x-5=0
x-8=0
x=8
b, x^2(x-1)-4x^2+8x-4=0
x^2(x-1)-(4x^2-8x+4)=0
x^2(x-1)-4(x^2-2x+1)=0
x^2(x-1)-4(x-1)^2=0
(x-1)(x^2-4)(x-1)=0
(x-1)(x-2)(x+2)(x-1)=0
=>x-1=0=>x=1
=>x-2=0=>x=2
=>x+2=0=>x=-2
=>x-1=0=>x=1
Vậy : x=1 ;x=2 và x=-2
c, (x-4)^2-36=0
(x-4)^2-6^2=0
(x-4-6)(x-4+6)=0
(x-10)(x+2)=0
=>x-10=0=>x=10
=>x+2=0=>x=-2
Vậy : x=10 và x=-2
k đúng cho mình nhé bạn !
b) Ta có: \(9x^4+8x^2-1=0\)
\(\Leftrightarrow9x^4+9x^2-x^2-1=0\)
\(\Leftrightarrow9x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(9x^2-1\right)=0\)
mà \(x^2+1>0\forall x\)
nên \(9x^2-1=0\)
\(\Leftrightarrow9x^2=1\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
a. \(x^2-25-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b. \(\left(3x+1\right)^2=\left(2x-5\right)\\ \Leftrightarrow9x^2+6x+1=2x-5\\ \Leftrightarrow9x^2+6x-2x=-5-1\\ \Leftrightarrow9x^2+4x=-6\\ \Leftrightarrow x\left(9x+4\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\9x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{10}{9}\end{matrix}\right.\)
c. \(2x^2-7x+6=0\\ \Leftrightarrow2x^2-7x=-6\\ \Leftrightarrow x\left(2x-7\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{2}\end{matrix}\right.\)
a, \(\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=-2;x=5\)
b, bạn ktra lại đề, thường thường ngta hay cho 2 vế cùng bình phương
c, \(2x^2-7x+6=0\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\Leftrightarrow x=\dfrac{3}{2};x=2\)