a) 

<=> \(x\left(0,2-1,2\right)+3,7=-6,3\)

<=> \(-x=-10\)

<=> \(x=10\)

b) 

<=> \(x\left(x-1\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

d) 

<=> \(2\sqrt{x+1}=8\)

<=> \(\sqrt{x+1}=4\)

<=> \(x=15\)

e) 

<=> \(\orbr{\begin{cases}1-x=\sqrt{2}-0,\left(1\right)\\1-x=0,\left(1\right)-\sqrt{2}\end{cases}}\)

<=> \(\orbr{\begin{cases}1+0,\left(1\right)-\sqrt{2}=x\\x=1+\sqrt{2}-0,\left(1\right)\end{cases}}\)

a) 0,2x + ( -1, 2 )x + 3, 7 = -6, 3

<=> x( 0,2 - 1, 2 ) + 3, 7 = -6, 3

<=> -x = -10

<=> x = 10

b) x² = x

<=> x² - x = 0

<=> x( x - 1 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

c) 0,(12) : 1,(6) = x : 0,(4)

<=> 4/33 : 5/3 = x : 4/9

<=> 4/55 = x : 4/9

<=> x = 16/495

d) \(2\sqrt{x+1}-3=5\)

\(\Leftrightarrow2\sqrt{x+1}=8\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\)

e) \(\left|1-x\right|=\sqrt{2}-0,\left(1\right)\)

\(\Leftrightarrow\left|1-x\right|=\sqrt{2}-\frac{1}{9}\)

\(\Leftrightarrow\left|1-x\right|=\frac{-1+9\sqrt{2}}{9}\)

\(\Leftrightarrow\orbr{\begin{cases}1-x=\frac{-1+9\sqrt{2}}{9}\\1-x=\frac{1-9\sqrt{2}}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{10-9\sqrt{2}}{9}\\x=\frac{8+9\sqrt{2}}{9}\end{cases}}\)

a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)

d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)

e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)

g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)

h)Tương tự các câu trên

i) x = 0

k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)

l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)

=> x + 1 = 3 => x = 2

x.(x+1)=0

suy ra x=0 hoac x+1=0

                               x=0-1

                              x=-1

vay x=0 hoac  x=-1

mấy câu sau cũng làm tương tự

Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)

           \(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)

          \(=100.\frac{2}{101}=\frac{200}{101}\)

d,\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\\ \Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\\ \Leftrightarrow x-\frac{2}{9}=\frac{4}{9}\\ \Leftrightarrow x=\frac{6}{9}\)

Vậy...

a) \(\left(x-3\right).\left(4-5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\4-5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+3\\5x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=4:5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{4}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{3;\frac{4}{5}\right\}.\)

b) \(\left|x+\frac{3}{4}\right|+\frac{1}{3}=0\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=0-\frac{1}{3}\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=-\frac{1}{3}.\)

Ta luôn có: \(\left|x\right|\ge0\) \(\forall x.\)

\(\Rightarrow\left|x+\frac{3}{4}\right|>-\frac{1}{3}\)

\(\Rightarrow\left|x+\frac{3}{4}\right|\ne-\frac{1}{3}.\)

Vậy \(x\in\varnothing.\)

c) \(5^x.\left(5^3\right)^2=625\)

\(\Rightarrow5^x.5^6=5^4\)

\(\Rightarrow5^{x+6}=5^4\)

\(\Rightarrow x+6=4\)

\(\Rightarrow x=4-6\)

\(\Rightarrow x=-2\)

Vậy \(x=-2.\)

Chúc bạn học tốt!

a,

\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)

b,

\(\left(5x-1\right)\left(2x+\dfrac{1}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

c,

\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)

\(\Rightarrow5\left|x+1\right|^2=180\)

\(\Rightarrow\left|x+1\right|^2=36\)

Mà \(\left|x+1\right|\ge0\)

=> x + 1 = 6 <=> x = 7

a, \(\left(x-1\right).\left(x+2\right)\)\(>0\Rightarrow\orbr{\begin{cases}x-1< 0;x+2< 0\left(loai\right)\Rightarrow x< 1\\x-1>0;x+2>0\Rightarrow x>1;x>-2\end{cases}}\)

=> -2 < x < 1

Câu b và câu d làm tương tự nha bạn(Câu b thì xét khác dấu) 

a) a=  2 và 1

b)    =      7

c=     5600 và 7899

d  5 và 6 

dễ thế mà ko biết làm ak!

1, x² = 0

=> x=0

2,x²=1

=> x= 1 hoặc x=-1

3,x²=3

=>\(x=\sqrt{3}\)

4,x²=6

=>\(x=\sqrt{6}\)

5,x²=7

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

a) (2 - x)(2x + 1) > 0

TH1:  \(\hept{\begin{cases}2-x>0\\2x+1>0\end{cases}\Rightarrow\hept{\begin{cases}x< 2\\x>-\frac{1}{2}\end{cases}\Rightarrow}-\frac{1}{2}< x< 2}\)

TH2: \(\hept{\begin{cases}2-x< 0\\2x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x>2\\x< -\frac{1}{2}\end{cases}\left(vl\right)}}\)(vô lí)

Vậy: -1/2 < x < 2

b) (2x+3)(x + 1) < 0

TH1: \(\hept{\begin{cases}2x+3>0\\x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-\frac{3}{2}\\x< -1\end{cases}\Rightarrow-\frac{3}{2}< x< -1}}\)

TH2: \(\hept{\begin{cases}2x+3< 0\\x+1>0\end{cases}\Rightarrow\hept{\begin{cases}\left(x< -\frac{3}{2}\right)\\x>-1\end{cases}}\left(vl\right)}\)(vô lí)

Vậy -3/2 < x < -1