a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)

b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)

c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)

\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)

\(\Rightarrow x=-2\)

d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{25}{9}\)

e) \(\dfrac{1}{2}x+650\%x-x=-6\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=\dfrac{-6}{6}=-1\)

g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)

\(\Rightarrow2x-1-3+x=2-x\)

\(\Rightarrow3x-4=2-x\)

\(\Rightarrow3x+x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

h) \(x-\dfrac{2}{20}=-\dfrac{5}{2}-x\)

\(\Rightarrow x+x=-\dfrac{5}{2}+\dfrac{2}{20}\)

\(\Rightarrow2x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}:2=-\dfrac{6}{5}\)

i) \(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\)

\(\Rightarrow\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

\(\Rightarrow\dfrac{x}{2}-1=\sqrt[3]{-\dfrac{27}{8}}\)

\(\Rightarrow\dfrac{x}{2}-1=-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{x}{2}=-\dfrac{3}{2}+1\)

\(\Rightarrow x=-\dfrac{1}{2}.2=-1\)

k) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

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

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

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

bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

cũng dễ thôi

a) \(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

 0,5 x + 0,6 ( x - 2 ) = 3

0,5 x + 0.6 x - 1,2 = 3

1,1 x = 4,2

x = \(\frac{42}{11}\)

Kết luận: 

b) \(\frac{1}{3}x-0,5x=0,75\)

\(\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)

\(-\frac{1}{6}x=\frac{3}{4}\)

\(x=-\frac{9}{2}\)

Kết luận: 

c) \(\frac{3}{-2}x-0,5x=75\%\)

-1,5x - 0,5x = 0,75 

-2x = 0,75 

x = -0,375

Kết luận: 

d) \(-\frac{2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)

-0,4 x + 0,25 = 0,75 - 0,75 x 

-0,4 x + 0,75 x = 0,75 - 0,25

0,35 x = 0,5 

x = \(\frac{10}{7}\)

Kết luận: 

\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)

\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)

\(\frac{11}{10}x-\frac{6}{5}=3\)

\(\frac{11}{10}x=\frac{21}{5}\)

\(x=\frac{42}{11}\)

\(b,\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)

\(\left(\frac{1}{3}-\frac{1}{2}\right)x=\frac{3}{4}\)

\(\frac{-1}{6}x=\frac{3}{4}\)

\(x=\frac{-9}{2}\)

\(c,\frac{3}{-2}x-\frac{1}{2}x=75\%\)

\(\left(\frac{3}{-2}-\frac{1}{2}\right)x=\frac{3}{4}\)

\(-2x=\frac{3}{4}\)

\(x=\frac{-3}{8}\)

\(\frac{-2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)

\(\frac{-2}{5}x+\frac{3}{4}x=\frac{3}{4}-\frac{1}{4}\)

\(\left(\frac{-2}{5}+\frac{3}{4}\right)x=\frac{1}{2}\)

\(\frac{7}{20}x=\frac{1}{2}\)

\(x=\frac{10}{7}\)

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 \(x=4;x=-8\)

Ta có S=1/2^2+1/3^2+1/4^2+...+1/9^2

<1/2²+1/2*3+1/3*4+....+1/8*9

=1/2²+1/2-1/3+1/3-1/4+....+1/8-1/9

=1/4+1/2-1/9=23/36<32/36=8/9 (♪)

Ta lại có S=1/2^2+1/3^2+1/4^2+...+1/9^2

>1/2²+1/3*4+1/4*5+....+1/9*10

=1/2²+1/3-1/4+1/4-1/5+........+1/9-1/10

=1/2²+1/3-1/10

=19/20>8/20=2/5 ( ♫)

Từ (♪)( ♫) cho ta đpcm 

3: 

\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)

4:

=>2/3:x=-2-1/3=-7/3

=>x=-2/3:7/3=-2/7

5:

AC=CB=12/2=6cm

IB=6/2=3cm