a)x=30

b)x=65

a) x-\(\left(\frac{50x}{100}+\frac{25x}{200}\right)\)=\(11\frac{1}{4}\)

<=>x - \(\frac{1}{2}x-\frac{1}{8}x\)=\(\frac{11.4+1}{4}\)

<=>\(\frac{3}{8}x=\frac{45}{4}\)

<=>x=\(\frac{45}{4}:\frac{3}{8}\)

<=>x=30

Vậy x=30

c) (x-5).\(\frac{30}{100}\)=\(\frac{20x}{100}+5\)

<=>(x-5).\(\frac{3}{10}\)=\(\frac{x}{5}\)+5

<=>\(\frac{3x}{10}-\frac{15}{10}=\frac{x}{5}+5\)

<=>\(\frac{3x}{10}-\frac{x}{5}=5+\frac{15}{10}\)

<=>\(\frac{x}{10}=\frac{13}{2}\)

<=>x=\(\frac{13.10}{2}\)

<=>x=65

Vậy: x=65

\(\left(x-5\right).\frac{30}{100}=\frac{20x}{100}+5\)

\(\Leftrightarrow\left(x-5\right).0,3=0,2.x+5\)

\(\Leftrightarrow0,3.x-1,5=0,2.x+5\)

\(\Leftrightarrow0,3.x-0,2.x=5+1,5\)

\(\Leftrightarrow0,1.x=6,5\)

\(\Leftrightarrow x=65\)

\(\frac{\left(x-5\right).30}{100}=\frac{20x}{100}+\frac{500}{100}\)

\(\frac{30x-150}{100}=\frac{20x+500}{100}\)

=> 30x - 150 = 20x + 500

30x - 20x = 500 + 150

10x = 650

x = 65

Ta có: \(\left(x-5\right)\frac{30}{100}=\frac{20x}{100}+5\)

\(\Rightarrow\left(x-5\right)\frac{3}{10}=\frac{x}{5}+5\)

\(\Rightarrow\frac{3x-15}{10}=\frac{25+x}{5}\)

\(\Rightarrow5\left(3x-15\right)=10\left(25+x\right)\)

\(\Rightarrow15x-75=250+10x\)

\(\Rightarrow15x-10x=250+75\)

\(\Rightarrow5x=325\Rightarrow x=\frac{325}{5}=65\)

\(\left(x-8\right).\frac{3}{10}=\frac{1}{5}x+5\)

\(\frac{3}{10}x-\frac{3}{2}=\frac{1}{5}x+5\)

\(\frac{3}{10}x=\frac{1}{5}x+6,5\)

\(\Rightarrow\frac{1}{10}x=6,5\)

\(x=\frac{6,5}{\frac{1}{10}}\)

\(x=65\)

Vậy \(x=65\)

1a Để \(\frac{x+1}{2}\)=\(\frac{8}{x+1}\)

\(\Rightarrow\)x+1.(x+1)=2.8=16

\(\Rightarrow\)x+1(x+1)=4.4

suy ra x+1=4

x=4-1

x=3

a)(x+1)(x+1)=16

(x+1)^2=4^2

+)x+1=4

x=3

+)x+1=-4

x=-5

Giải phương tình nha :v 

a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow2160-198x=160x+12\)

\(\Leftrightarrow358x=2148\)

\(\Leftrightarrow x=6\)

Vậy nghiệm của pt x=6

b)  \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)

\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)

\(\Leftrightarrow-77x-21=48x-396\)

\(\Leftrightarrow125x=375\)

\(\Leftrightarrow3\)

Vậy nghiệm của pt x=3

c)\(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)

\(\Leftrightarrow\frac{15\left(x-3\right)}{20}+\frac{8x-21}{20}=\frac{3x+3}{5}+\frac{30}{5}\)

\(\Leftrightarrow\frac{23x-66}{20}=\frac{3x+33}{5}\)

\(\Leftrightarrow115x-330=60x+660\)

\(\Leftrightarrow55x=990\)

\(\Leftrightarrow x=18\)

Vậy nghiệm của pt x=18

d) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2\left(3x-1\right)}{5}-\frac{3x+2}{10}\)

\(\Leftrightarrow\frac{6x+3}{4}-\frac{20}{4}=\frac{4\left(3x-1\right)}{10}-\frac{3x+2}{10}\)

\(\Leftrightarrow\frac{6x-17}{4}=\frac{9x-6}{10}\)

\(\Leftrightarrow60x-170=36x-24\)

\(\Leftrightarrow24x=146\)

\(\Leftrightarrow x=\frac{73}{12}\)

Vậy nghiệm của pt \(x=\frac{73}{12}\)

a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)

\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)

=>-9/10=-9/10(luôn đúng)

b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)

=>347x+780=1552

=>347x=772

hay x=772/347

a/ Ta có 

\(K^4+\frac{1}{4}=K^4+K^2+\frac{1}{4}-K^2=\left(K^2+\frac{1}{2}\right)^2-K^2=\left(K^2+K+\frac{1}{2}\right)\left(K^2-K+\frac{1}{2}\right)\)

Ta lại có 

\(K^2+K+\frac{1}{2}=\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\)

\(\Rightarrow K^4+\frac{1}{4}=\left(K^2-K+\frac{1}{2}\right)\left(\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\right)\)

Áp dụng vào bài toán ta được

\(=\frac{101^2-101+0,5}{1^2-1+0,5}=20201\)\(1S=\frac{\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)\left(5^2-5+0,5\right)...\left(100^2-100+0,5\right)\left(101^2-101+0,5\right)}{\left(1^2-1+0,5\right)\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)...\left(99^2-99+0,5\right)\left(100^2-100+0,5\right)}\)

b/

\(\frac{3\left(x+y\right)}{3\sqrt{x\left(4x+5y\right)}+3\sqrt{y\left(4y+5x\right)}}\)

\(\ge\frac{3\left(x+y\right)}{\frac{9x+4x+5y}{2}+\frac{9y+4y+5x}{2}}\)

\(=\frac{1}{3}\)

Dấu = xảy ra khi x = y

Bấm sao mà nói đẩy đáp số lên trên mất rồi

\(\Rightarrow1S=\frac{101^2-101+0,5}{1^2-1+0,5}=20201\)