a) \(\sqrt{x^4}=2\)( ĐK x ∈ R )

⇔ \(\sqrt{\left(x^2\right)^2}=2\)

⇔ \(\left|x^2\right|=2\)

⇔ \(\orbr{\begin{cases}x^2=2\\x^2=-2\left(loai\right)\end{cases}}\)

⇔ x² - 2 = 0

⇔ ( x - √2 )( x + √2 ) = 0

⇔ x - √2 = 0 hoặc x + √2 = 0

⇔ x = ±√2 

b) \(3\sqrt{x+1}-8=0\)( ĐK x ≥ -1 )

⇔ \(3\sqrt{x+1}=8\)

⇔ \(\sqrt{x+1}=\frac{8}{3}\)

⇔ \(x+1=\frac{64}{9}\)

⇔ \(x=\frac{55}{9}\)( tm )

c) \(2\sqrt{x-3}+\sqrt{25x-75}=14\)( ĐK x ≥ 3 ) ( Vầy hợp lí hơn á )

⇔ \(2\sqrt{x-3}+\sqrt{5^2\left(x-3\right)}=14\)

⇔ \(2\sqrt{x-3}+5\sqrt{x-3}=14\)

⇔ \(7\sqrt{x-3}=14\)

⇔ \(\sqrt{x-3}=2\)

⇔ \(x-3=4\)

⇔ \(x=7\)( tm )

d) \(\sqrt{\left(3x-1\right)^2}=5\)( ĐK x ∈ R )

⇔ \(\left|3x-1\right|=5\)

⇔ \(\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)

e) \(\sqrt{x^2+4x+4}-6=0\)( ĐK x ∈ R )

⇔ \(\sqrt{\left(x+2\right)^2}=6\)

⇔ \(\left|x+2\right|=6\)

⇔ \(\orbr{\begin{cases}x+2=6\\x+2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)

\(a)\)\(\sqrt{x^4}=2\)\(\Leftrightarrow\)\(x^2=2\)\(\Rightarrow\)\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)

Vậy \(x=\sqrt{2}\)\(hoặc\)\(x=-\sqrt{2}\)

​\(b)\)\(ĐK:x\ge0\)

\(3\sqrt{x+1}-8=0\)\(\Leftrightarrow\)\(3\sqrt{x}=8\)\(\Leftrightarrow\)\(\sqrt{x}=\frac{8}{3}\)\(\Leftrightarrow\)\(x=(\frac{8}{3})^2\)\(\Leftrightarrow\)\(x=\frac{64}{9}\)\((TM)\)

Vậy \(x=\frac{64}{9}\)

\(d)\)\(\sqrt{(3x-1)^2}=5\)\(\Leftrightarrow\)\(|3x-1|=5\)\((1)\)

Vậy \(x\in\hept{2;\frac{-4}{3}}\)

                -Nếu \(x\ge-2\)thì \(\left(2\right)\Leftrightarrow x+2=6\Leftrightarrow x=4(TM)\)

                -Nếu \(x< -2\)thì \(\left(2\right)\Leftrightarrow-\left(x+2\right)=6\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(TM\right)\)

Vậy \(x=4;x=-8\)

a) Ta có: \(3-\left(17-x\right)=-12\)

\(\Leftrightarrow3-17+x+12=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: x=2

b) Ta có: \(\left(2x+4\right)\left(10-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\10-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;5\right\}\)c) Ta có: \(\left|x-9\right|=-2+17\)

\(\Leftrightarrow\left|x-9\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}x-9=15\\x-9=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{24;-6\right\}\)

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

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

\(x\cdot\dfrac{-1}{14}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{-1}{14}\)

\(x=\dfrac{-42}{5}\)

Bài 2 : a, x = -36/9 = -4

b, đề sai 

c, <=> -2 =< x =< -3 => x = -1 

Bài 1: 

a: 2/8=9/36; 2/9=8/36; 8/2=36/9; 9/2=36/8

b: -2/4=9/-18; -2/9=4/-18; 4/-2=-18/9; 9/-2=-18/4

Bài 2: 

a: =>x/3=-4/3

hay x=-4

Câu b đề sai rồi bạn

1/2+4/3*1/6

=1/2+4/18

=13/18

1/2 + 4/3 = 3/6 + 8/6 = 11/6

4/5 + 3/4 : 2/3 = 4/5 + 3/4 * 3/2 = 4/5 + 9/8 = 32/40 + 45/40 = 77/40

7/2 - 4/3 * 1/6 = 7/2 - 2/9 = 63/18 - 4/18 = 59/18

4/5 * 4/7 : 2/3 = 16/35 : 2/3 = 16/35 * 3/2 = 24/35

1/4 : 5/3 + 1/6 = 1/4 * 3/5 + 1/6 = 3/20 + 1/6 = 9/60 + 10/60 = 19/60

3 - 4/5 : 1/3 = 3 - 4/5 * 3/1 = 3 - 12/5 = 15/5 - 12/5 = 3/5

5/9 * 2/7 - 3/7 = 10/63 - 3/7 = 10/63 - 27/63 = -17/63  

a) \(15\times\frac{2121}{4343}+15\times\frac{222222}{434343}=15\times\frac{21}{43}+15\times\frac{22}{43}=15\times\left(\frac{21}{43}+\frac{22}{43}\right)=15\times1=15\)

b) \(\frac{16\times25+44\times100}{29\times96+142\times48}=\frac{25\times\left(16+44\times4\right)}{48\times\left(29\times2+142\right)}=\frac{25\times192}{48\times200}=\frac{4}{8}=\frac{1}{2}\)

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Đặt \(A=2^{x+9}-2^{x+8}-2^{x+7}-...-2^{x+1}-2^x\)

\(\Rightarrow2A=2\left(2^{x+9}-2^{x+8}-...-2^x\right)\)

\(\Rightarrow2A=2^{x+9}.2^1-2^{x+8}.2^1-...-2^x.2^1\)

\(\Rightarrow2A=2^{x+10}-2^{x+9}-...-2^{x+1}\)

\(\Rightarrow A=2A-A=2^{x+10}-2^{x+9}-...-2^{x+1}-\left(2^{x+9}-2^{x+8}-...-2^{x+1}-2^x\right)=2^{x+10}-2^{x+9}-2^{x+9}+2^x\)

\(\Rightarrow A=2^{x+10}-2.2^{x+9}+2^x=2^{x+10}-2^{x+10}+2^x=2^x\)

\(\Rightarrow2^x=1024\Rightarrow x=10\)