giúp mik vs cần luôn
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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)\(\left(\sqrt{\frac{4}{9}}+\sqrt{\frac{25}{144}}-\sqrt{\frac{49}{81}}\right):\sqrt{\frac{121}{36}}\)
\(=\left(\frac{\text{2}}{3}+\frac{5}{12}-\frac{7}{9}\right):\frac{11}{6}\)
\(=\frac{11}{36}:\frac{11}{6}=\frac{1}{6}\)
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\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
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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) \(\left(-6,5\right).2,8+2,8.\left(-3,5\right)=2,8\left(-10\right)=-28\)
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\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{10}=\frac{y}{15}\\\frac{z}{4}=\frac{y}{5}\Leftrightarrow\frac{z}{12}=\frac{y}{15}\end{cases}\Leftrightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12};x+y-z=-39}\)
Tính chất dãy tỉ số bằng nhau:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y-z}{10+15-12}=-\frac{39}{13}=-3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-3\Leftrightarrow x=-30\\\frac{y}{15}=-3\Leftrightarrow y=-45\\\frac{z}{12}=-3\Leftrightarrow z=-36\end{cases}}\)
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\(\left(x-1\right)^{x-2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x-2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x-2}\left[\left(x-1\right)^8-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^{x-2}=0\\\left(x-1\right)^8-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-1=\pm1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2;x=0\end{cases}}\)
Thử lại \(x=2,x=0\)thỏa mãn.
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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
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đề bài sai nha
BC=BC
suy ra ABC=ADC
vì ABC DÀI BẰNG ADC