\(\Leftrightarrow\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\z=\frac{-7}{20}\end{cases}}\)

\(\frac{2}{9}+\frac{1}{6}:x=\frac{5}{9}\)

=>\(\frac{1}{6}:x=\frac{5}{9}-\frac{2}{9}\)

=>\(\frac{1}{6}:x=\frac{1}{3}\)

=>\(x=\frac{1}{6}:\frac{1}{3}\)

=>\(x=\frac{1}{2}\)

B1:

Từ \(b=\frac{a+c}{2}\Rightarrow2b=a+c\left(1\right)\)

Từ \(c=\frac{2bd}{b+a}\)thay vào (1) ta được:

\(2b=a+\frac{2bd}{b+a}\)

\(\Leftrightarrow2b\left(b+a\right)=a\left(b+a\right)+2bd\)

\(\Leftrightarrow2b^2+2ab=ab+a^2+2bd\)

\(\Leftrightarrow2b^2+ab-a^2-2bd=0\)

\(\Leftrightarrow2b\left(b-d\right)+a\left(b-a\right)=0\)

\(\Leftrightarrow2b\left(b-d\right)=a\left(a-b\right)\Leftrightarrow\frac{2b}{a}=\frac{a-b}{b-d}\)

B2: Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}hay2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

Do đó: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)

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

TH1: Nếu \(x-\frac{2}{3}\ge0\Rightarrow x\ge\frac{2}{3}\Rightarrow\left|x-\frac{2}{3}\right|=x-\frac{2}{3}\)

\(A=x+\frac{1}{2}-x+\frac{2}{3}=\frac{7}{6}\left(1\right)\)

TH2: Nếu \(x-\frac{2}{3}< 0\Rightarrow x< \frac{2}{3}\Rightarrow\left|x-\frac{2}{3}\right|=-x+\frac{2}{3}\)

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

Vì \(x< \frac{2}{3}\Rightarrow2x-\frac{1}{6}< \frac{7}{6}\left(2\right)\)

Từ (1) và (2) => GTLN của A là \(\frac{7}{6}\)khi \(x\ge\frac{2}{3}\)

i donnot

1.

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

Ta có\(x-2y+3z=22\)

\(\Leftrightarrow2k-6k+15k=22\)

\(\Leftrightarrow11k=22\Leftrightarrow k=2\)

Do đó  \(\hept{\begin{cases}\frac{x}{2}=2\Leftrightarrow x=4\\\frac{y}{3}=2\Leftrightarrow y=6\\\frac{z}{5}=2\Leftrightarrow z=10\end{cases}}\)

2.

Theo tính chất dãy tỉ số bằng nhau\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+y^2-z^2}{4+9-25}=\frac{150}{-12}=-\frac{25}{2}\)

Ta có

 \(\frac{x}{2}=-\frac{25}{2}\Leftrightarrow x=2.\left(-25\right):2=-25\)

\(\frac{y}{3}=-\frac{25}{2}\Leftrightarrow y=3.\left(-25\right):2=-\frac{75}{2}\)

\(\frac{z}{5}=-\frac{25}{2}\Leftrightarrow z=5.\left(-25\right):2=-\frac{125}{2}\)

Thử lại ko đúng cách đặt thì \(k^2=-\frac{25}{2}\left(ktm\right)\) mình nghĩ đề sai

x² + 4 = 2³

x² = 8 - 4

x² = 4

x² = 2²

x = 2

\(x^2+4=2^3\Leftrightarrow x^2=8-4=4\)

\(\Rightarrow x=\pm2\)