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

\(x-\frac{9}{4}=\frac{7}{2}\)

\(x=\frac{7}{2}+\frac{9}{4}\)

\(x=\frac{14}{4}+\frac{9}{4}\)

\(x=\frac{23}{4}\)

Ta có: \(\frac{x}{2}=\frac{15}{20}\Leftrightarrow\frac{10x}{20}=\frac{15}{20}\Leftrightarrow10x=15\Leftrightarrow x=\frac{3}{2}\)

Vậy \(x=\frac{3}{2}\)

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

x.4=2.3

x.4=6

x   =6:4

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

\(\Rightarrow\)x=3

Ta có : \(\frac{15}{20}\)\(=\)\(\frac{15:5}{20:5}\)\(=\)\(\frac{3}{4}\)

Vì  \(\frac{x}{2}\)\(=\)\(\frac{15}{20}\)\(\Rightarrow\)\(\frac{x}{2}\)\(=\)\(\frac{3}{4}\) 

\(\Rightarrow\)x . 4  \(=\)3 . 2 \(\Rightarrow\)x.4  \(=\)6

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

Vậy   x \(=\)\(\frac{3}{2}\)

\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\)

\(=\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x+2\right)^2-\left(x-2\right)^2-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+4x+4-x^2+4x-4-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8x-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8}{x+2}\)

bạn ơi, cho mình hỏi

4-x² thì sẽ = 2²-x² = (2-x)(2+x)

vậy sao bạn ghi là (x-2)(x+2) vậy???

Biến đổi như thế này nha bạn :

\(\frac{16}{4-x^2}=\frac{-1\cdot16}{-1\cdot\left(4-x^2\right)}\)

\(=\frac{-16}{-4+x^2}=\frac{-16}{x^2-4}\)

\(=\frac{-16}{\left(x-2\right)\left(x+2\right)}\)

1/ Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{2^2}{4}=\frac{4}{4}=1\)

Dấu "=" xảy ra khi x=y=1

Máy mình bị lỗi nên ko nhìn được các bài tiếp theo

Chúc bạn học tốt :)

Ta có : x+y=2 => x=2-y. Thay vào bt ta đc : xy= (2-y).y = 2y -y^2    

Vì y^2 >= 0 =>2y-y^2 nhỏ hơn hoặc bằng 0

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

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

\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)

\(\left(2x-1\right)=-\frac{4}{63}\)

2x= -4/63 + 1

2x = 59/63

x = 59/63 : 2

x = 59/126

1/3:(2.x-1)=-5-1/4

1/3:(2.x-1)=-21/4

2.x-1=1/3:-21/4

2.x-1=-4/63

2.x=-4/63+1

2.x=\(3\frac{59}{63}\)

x=\(3\frac{59}{63}\):2

x=\(1\frac{61}{63}\)

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

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

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

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-21}{4}\)

\(2x-1=\frac{1}{3}\div\frac{-21}{4}\)

\(2x-1=\frac{1}{3}.\frac{-4}{21}\)

\(2x-1=\frac{-4}{63}\)

\(2x=\frac{-4}{63}+1\)

\(2x=\frac{-4}{63}+\frac{63}{63}\)

\(2x=\frac{59}{63}\)

\(x=\frac{59}{63}\div2\)

\(x=\frac{59}{63}.\frac{1}{2}\)

\(x=\frac{59}{126}\)

\(\frac{1}{4}:\frac{x}{9}=\frac{1}{4}\)

\(\frac{x}{9}=\frac{1}{4}:\frac{1}{4}\)

\(\frac{x}{9}=1\)

\(\Rightarrow x=9\)

Trl:

Bn Vũ Huyền lm đúng r nha!

Hok tốt!!

\(\frac{1}{4}:\frac{x}{9}=\frac{1}{4}\)

\(\Rightarrow\frac{x}{9}=\frac{1}{4}:\frac{1}{4}=1\)

\(\Rightarrow x=9\)

~học tốt~

\(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

\(\Rightarrow\frac{5}{x}=\frac{1}{8}-\frac{y}{4}\)

\(\Rightarrow\frac{5}{x}=\frac{1-2y}{8}\)

\(\Rightarrow x\left(1-2y\right)=40\)

tu xet bang

tớ có cách khác:))

\(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

\(\Rightarrow\frac{20+xy}{4x}=\frac{1}{8}\)

\(\Rightarrow\frac{40+2xy}{8x}=\frac{x}{8x}\)

\(\Rightarrow40+2xy=x\)

\(\Rightarrow40=x\left(1-2y\right)\)

Cách này xem cho vui nha.dài hơn cách của Phương Uyên.

đề học sinh giỏi hồi chiều ak!!!!!!!!! khó v:

a, => |5/3.x| = 1/6

=> 5/3.x = -1/6 hoặc 5/3.x = 1/6

=> x = -1/10 hoặc x = 1/10

Tk mk nha