Ai trả lời nhanh mk k cho

Fat you

hihihihi

a)  \(\left|\sqrt{2}-x\right|=\sqrt{2}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{2}-x=\sqrt{2}\\\sqrt{2}-x=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\sqrt{2}\end{cases}}}\)

b) \(\left|x+1\right|=\sqrt{3}+2\)

\(\Rightarrow\orbr{\begin{cases}x+1=\sqrt{3}+2\\x+1=-\sqrt{3}-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}+1\\x=-\sqrt{3}-3\end{cases}}\)

\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)

\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)

\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)

\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)

\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)

\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)

\(3^x+\frac{4}{9}=9+\frac{4}{9}\)

\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

a: \(-12:\left(\frac34-\frac56\right)^2\)

\(=-12:\left(\frac{9}{12}-\frac{10}{12}\right)^2\)

\(=-12:\left(-\frac{1}{12}\right)^2=-12\cdot144=-1728\)

b: \(10\cdot\sqrt{0,01}\cdot\sqrt{\frac{16}{9}}+3\sqrt{49}-\frac16\cdot\sqrt4\)

\(=10\cdot0,1\cdot\frac43+3\cdot7-\frac16\cdot2\)

\(=\frac43+21-\frac13=21+1=22\)

c: Đặt \(\frac{x}{6}=\frac{y}{3}=\frac{z}{2}=k\)

=>x=6k; y=3k; z=2k

x-2y+4z=8

=>\(6k-2\cdot3k+4\cdot2k=8\)

=>8k=8

=>k=1

=>\(\begin{cases}x=6\cdot1=6\\ y=3\cdot1=3\\ z=2\cdot1=2\end{cases}\)

d: \(\left|\frac14+x\right|-\frac13=\frac25\)

=>\(\left|x+\frac14\right|=\frac25+\frac13=\frac{11}{15}\)

=>\(\left[\begin{array}{l}x+\frac14=\frac{11}{15}\\ x+\frac14=-\frac{11}{15}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{15}-\frac14=\frac{44-15}{60}=\frac{29}{60}\\ x=-\frac{11}{15}-\frac14=\frac{-44-15}{60}=-\frac{59}{60}\end{array}\right.\)

Giả theo cách lớp 7 nha:

Đặt \(\hept{\begin{cases}\sqrt{6-x}=a\\\sqrt{x+2}=b\end{cases}}\)

\(\Rightarrow a^2+b^2=8\)

Ta có:

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

\(\Leftrightarrow2ab\le a^2+b^2\)

\(\Leftrightarrow a^2+b^2+2ab\le2\left(a^2+b^2\right)\)

\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)=2\cdot8=16\)

\(\Leftrightarrow a+b\le4\)

Dấu = xảy ra khi \(a=b=2\)

\(\Leftrightarrow x=2\)

\(ĐKXĐ:-2\le x\le6\)

Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\le\sqrt{2.\left(a+b\right)}\) với \(a,b\ge0\) ta có :

\(y=\sqrt{6-x}+\sqrt{x+2}\le\sqrt{2.\left(6-x+x+2\right)}=\sqrt{2.8}=4\)

Dấu "=" xảy ra \(\Leftrightarrow6-x=x+2\Leftrightarrow x=2\)

Vậy \(y_{min}=4\) khi \(x=2\)