2*x^2+0.82=1
7-√x=0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)7-\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}=7-0\)
\(\Rightarrow\sqrt{x}=7\)
Vậy \(x=7\)
\(b)4^{x^2}-1=0\)
\(\Rightarrow4^{x^2}=0+1\)
\(\Rightarrow4^{x^2}=1\)
\(\Rightarrow x^2=\dfrac{1}{4}\)
\(\Rightarrow x=\pm\sqrt{\dfrac{1}{4}}=\pm\dfrac{1}{2}\)
Vậy ..................
\(c)2^{x^2}+0,82=1\)
\(\Rightarrow2^{x^2}+0=1\)
\(\Rightarrow2^{x^2}=1\)
\(\Rightarrow x^2=\dfrac{1}{2}\)
\(\Rightarrow x=\pm\sqrt{\dfrac{1}{2}}\)
Vậy ......................
Chúc bạn học tốt!
*) \(4x^2-1=0\)
\(\Rightarrow4x^2=1\Rightarrow x^2=\dfrac{1}{4}\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
*) \(2x^2+0,82=1\)
\(\Rightarrow2x^2=1-0,82=\dfrac{9}{50}\)
\(\Rightarrow x^2=\dfrac{9}{100}\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{3}{10}\end{matrix}\right.\)
*) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{12}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Giải:
a) \(4x^2-1=0\)
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
b) \(2x^2+0,82=1\)
\(\Leftrightarrow2x^2=0,18\)
\(\Leftrightarrow x^2=0,09\)
\(\Leftrightarrow x=\pm0,3\)
Vậy ...
c) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{12}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
\(a.\)
\(2x^2+0.82=1\)
\(\Rightarrow2x^2+0=1\)
\(\Rightarrow2x^2=1\)
\(\Rightarrow x^2=\dfrac{1}{2}\)
\(\Rightarrow x=\pm\sqrt{\dfrac{1}{2}}\)
\(b.\)
\(4x^2-1=0\)
\(4x^2=1\)
\(\Rightarrow x^2=\dfrac{1}{4}\)
\(\Rightarrow x=\pm\sqrt{\dfrac{1}{4}}=\pm\dfrac{1}{2}\)
\(c.\)
\(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{1}{4}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{12}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Lời giải:
a. $x^2-4x-5=0$
$\Leftrightarrow (x+1)(x-5)=0$
$\Leftrightarrow x+1=0$ hoặc $x-5=0$
$\Leftrightarrow x=-1$ hoặc $x=5$
b.
$5x^2-9x-2=0$
$\Leftrightarrow (x-2)(5x+1)=0$
$\Leftrightarrow x-2=0$ hoặc $5x+1=0$
$\Leftrightarrow x=2$ hoặc $x=\frac{-1}{5}$
c.
$(x^2+1)-5(x^2+1)+6=0$
$\Leftrightarrow a^2-5a+6=0$ (đặt $x^2+1=a$)
$\Leftrightarrow (a-2)(a-3)=0$
$\Leftrightarrow a-2=0$ hoặc $a-3=0$
$\Leftrightarrow x^2-1=0$ hoặc $x^2-2=0$
$\Leftrightarrow (x-1)(x+1)=0$ hoặc $(x-\sqrt{2})(x+\sqrt{2})=0$
$\Leftrightarrow x\in\left\{\pm 1; \pm \sqrt{2}\right\}$
d.
$(x^2+6x)-2(x+3)^2-17=0$
$\Leftrightarrow (x^2+6x+9)-2(x+3)^2-26=0$
$\Leftrightarrow (x+3)^2-2(x+3)^2-26=0$
$\Leftrightarrow -(x+3)^2-26=0$
$\Leftrightarrow (x+3)^2=-26<0$ (vô lý)
Do đó không tồn tại $x$ thỏa mãn.
a. ( x - 1200 )^2 + ( x + 2015 )^ 4 = 0
b. / 17 - x / . / y - 18 / = 0
c. / 17 - x / + / 17 - y / = 0
\(2x\left(x-17\right)+\left(17-x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-17\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=17\end{cases}}\)
a)
(2x-4)(x-22)=0
<=>2x-4=0 hoặc x-22=0
<=>x=2 hoặc x=22
b
(x-17)(x^2-16)=0
<=>x-17 =0 hoặc x^2-16=0
<=>x=17 hoặc x=4 hoặc x=-4
c
(x^2+3)(x+8)=0
Vì x^2+3>0
=>x+8=0
<=>x=-8
a) \(2.x^2+0,82=1\)
\(\Leftrightarrow2x^2=0,18\)
\(\Leftrightarrow x^2=0,09\)
\(\Leftrightarrow x=\sqrt{0,09}=\pm0,3\)
vậy pt có tập nghiệm x={0,3;-0,3}
b) \(7-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}=7\)
\(\Leftrightarrow x=7^2=49\)
vậy x=49