1) \(\sqrt{25-9}=\sqrt{16}\) \(=\pm4\)

2) \(\sqrt{0,01}-\sqrt{0,25}\)

\(=\sqrt{\frac{1}{100}}-\sqrt{\frac{25}{100}}\)

\(=\frac{1}{10}-\frac{5}{10}\)

\(=\frac{-4}{10}\)

3)\(\left(\sqrt{\frac{4}{9}}+\sqrt{\frac{25}{144}}-\sqrt{\frac{49}{81}}\right):\sqrt{\frac{121}{36}}\)

\(=\left(\frac{\text{2}}{3}+\frac{5}{12}-\frac{7}{9}\right):\frac{11}{6}\)

\(=\frac{11}{36}:\frac{11}{6}=\frac{1}{6}\)

\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}\)

\(\sqrt{4+36+81}\)

\(=\sqrt{121}\)

\(=\pm11\)

a ) \(6,3+\left(-3,7\right)+2,4+\left(-0,3\right)=8,7-4=4,7\)

b) \(\left(-4,9\right)+5,5+4,9+\left(-5,5\right)=0\)

c) \(2,9+3,7+\left(-4,2\right)+\left(-2,9\right)+4,2=3,7\)

d) \(\left(-6,5\right).2,8+2,8.\left(-3,5\right)=2,8\left(-10\right)=-28\)

chơi free fire ko

\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{10}=\frac{y}{15}\\\frac{z}{4}=\frac{y}{5}\Leftrightarrow\frac{z}{12}=\frac{y}{15}\end{cases}\Leftrightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12};x+y-z=-39}\)

Tính chất dãy tỉ số bằng nhau:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y-z}{10+15-12}=-\frac{39}{13}=-3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-3\Leftrightarrow x=-30\\\frac{y}{15}=-3\Leftrightarrow y=-45\\\frac{z}{12}=-3\Leftrightarrow z=-36\end{cases}}\)

Bằng - 45

\(\left(x-1\right)^{x-2}=\left(x-1\right)^{x+6}\)

\(\Leftrightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x-2}=0\)

\(\Leftrightarrow\left(x-1\right)^{x-2}\left[\left(x-1\right)^8-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^{x-2}=0\\\left(x-1\right)^8-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-1=\pm1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2;x=0\end{cases}}\)

Thử lại \(x=2,x=0\)thỏa mãn. 

ta có x/2=y/3 suy ra x/10=y/15 (1)

z/4=y/5 suy ra z/12=y/15 (2)

từ (1),(2) suy ra x/10= y/15= z/12 áp dụng tính chất của dãy tỉ số bằng nhau có:

x/10=y/15=z/12 = x+y-z/10+15-12= -39/13=-2

x/10=-2 suy ra x=-2.10=-20

y/15=-2 suy ra y=-2.15=-30

z/12=-2 suy ra z=-2.12=-24

tìm x, y và z nhé

Ta có: 

\(\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\ge\left|x+2+1-x\right|=3\)

Dấu \(=\)khi \(\left(x+2\right)\left(1-x\right)\ge0\Leftrightarrow-2\le x\le1\).

Do đó \(\left|x+2\right|+\left|3x-1\right|+\left|x-1\right|=3\)

\(\Leftrightarrow\hept{\begin{cases}-2\le x\le1\\3x-1=0\end{cases}}\Leftrightarrow x=\frac{1}{3}\).