\(a)7-\sqrt{x}=0\)

\(\Rightarrow\sqrt{x}=7-0\)

\(\Rightarrow\sqrt{x}=7\)

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!

- V~V cả toán lớp 10

\(a.\)

\(2x^2+0.82=1\)

\(\Rightarrow2x^2+0=1\)

\(\Rightarrow2x^2=1\)

\(\Rightarrow x^2=\dfrac{1}{2}\)

\(\Rightarrow x=\pm\sqrt{\dfrac{1}{2}}\)

\(b.\)

\(4x^2-1=0\)

\(4x^2=1\)

\(\Rightarrow x^2=\dfrac{1}{4}\)

\(\Rightarrow x=\pm\sqrt{\dfrac{1}{4}}=\pm\dfrac{1}{2}\)

\(c.\)

\(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{1}{4}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{12}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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.

\(2x\left(x-17\right)+\left(17-x\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-17\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=17\end{cases}}\)

\(9x^2-18x=0\)

\(\Leftrightarrow9x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b) \(x\left(x-2\right)+5\left(2+x\right)=0\)

\(\Leftrightarrow x^2-2x+10+5x=0\)

\(\Leftrightarrow x^2+3x+10=0\)

Dễ thấy phương trình vô nghiệm do vế trái luôn dương

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