(a + 2c)(b + d) = (a + c)(b + 2d)

a(b + d) + 2c(b + d) = a(b + 2d) + c(b + 2d)

ab + ad + 2cb + 2cd = ab + 2ad + cb + 2cd

ab + ad + 2cb = ab + 2ad + cb

ad + 2cb = 2ad + cb

2ad - ad = 2cd - cd

ad = cb 

\(\frac{a}{b}=\frac{c}{d}\)

\(A=\frac{14-x}{4-x}=\frac{10+\left(4-x\right)}{4-x}=1+\frac{10}{4-x}\)

Để \(A=1+\frac{10}{4-x}\) đạt GTLN <=> \(4-x\) là số nguyên dương nhỏ nhất

=> \(4-x=1\Rightarrow x=3\)

Vậy \(A_{max}=1+\frac{10}{4-3}=11\)

Vật GTLN của A là 11 tại x = 3

x=3 nha!

Tk minh nhé!

\(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow a^2+2ab+b^2\ge2ab+2ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\) (đpcm)

Ta có : với a,b>0 theo bđt Cô si: a+b\(\ge\)\(2\sqrt{ab}\) 

=> (a+b)\(^2\)\(\ge\)4ab

 nhớ k mình nha ^^

\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)

thử suy nghĩ nào

a. neither do

either

b. too

c. , too

d. So was

e. too

Chắc vậy

a)either do/too

b)too

d)so was

e)too

Đặt \(\frac{3\left|x\right|+5}{3}=\frac{3\left|y\right|-1}{5}=\frac{3-z}{7}=k\)

\(\Rightarrow\left|x\right|=\frac{3k-5}{3}\Rightarrow2\left|x\right|=\frac{6k-10}{3}\)

\(\Rightarrow\left|y\right|=\frac{5k+1}{3}\Rightarrow7\left|y\right|=\frac{35k+7}{3}\)

\(\Rightarrow z=3-7k\Rightarrow3z=9-21k\)

Vì \(2\left|x\right|+7\left|y\right|+3z=-14\)\(\Rightarrow\frac{6k-10}{3}+\frac{35k+7}{3}+\left(9-21k\right)=-14\)

\(\Rightarrow\frac{\left(6k-10\right)+\left(35k+7\right)+\left(27-63k\right)}{3}=-14\)

\(\Rightarrow\frac{-22k+24}{3}=-14\)

\(\Rightarrow-22k+24=-42\)

\(\Rightarrow k=\frac{-42-24}{22}=3\)

\(\Rightarrow\left|x\right|=\frac{3.3-5}{3}=\frac{4}{3}\Rightarrow x=-\frac{4}{3};\frac{4}{3}\)

\(\Rightarrow\left|y\right|=\frac{5.3+1}{3}=\frac{16}{3}\Rightarrow y=-\frac{16}{3};\frac{16}{3}\)

\(\Rightarrow z=3-7.3=-18\)