Question

Tìm x,y,z

a) \(\sqrt{3-x}=5\)

b) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{81}{64}\)

c) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)và \(x^2-2y^2+z^2=44\)

d) \(\left(x-0,2\right)^{70}+\left(y+3,1\right)^{20}=0\)

Answer 1

mình chỉ biết làm 2 câu b and c thôi bạn thông cảm nha

Tìm x,y,z

b,\(\left(x+\frac{1}{2}\right)^2=\frac{81}{64}\)

Có \(\frac{81}{64}=\left(\frac{9}{8}\right)^2hoặc\frac{81}{64}=\left(-\frac{9}{8}\right)^2\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{9}{8}\right)^2hoặc\left(x+\frac{1}{2}\right)^2=\left(-\frac{9}{8}\right)^2\)

+TH1: \(\left(x+\frac{1}{2}\right)^2=\left(\frac{9}{8}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\frac{9}{8}\)

\(x=\frac{9}{8}-\frac{1}{2}\)

\(x=\frac{9-4}{8}\)

\(x=\frac{5}{8}\)

+TH2:\(\left(x+\frac{1}{2}\right)^2=\left(-\frac{9}{8}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=-\frac{9}{8}\)

\(x=-\frac{9}{8}-\frac{1}{2}\)

\(x=\frac{-9-4}{8}\)

\(x=\frac{-13}{8}\)

Vậy x= \(\frac{5}{8}\)hoặc x=\(\frac{-13}{8}\)

c, \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\) và \(x^2-2y^2+z^2\)

Ta có : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)\(\Leftrightarrow\)\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\Rightarrow\frac{x^2}{4}=\frac{2y^2}{18}=\frac{z^2}{25}\)

- Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x^2}{4}=\frac{2y^2}{18}=\frac{z^2}{25}=\frac{x^2-2y^2+z^2}{4-18+25}=\frac{44}{11}=4\)

- Do đó :

\(\frac{x^2}{4}=4\Leftrightarrow\frac{x}{2}=4\Rightarrow x=4.2=8\)

\(\frac{2y^2}{18}=4\Leftrightarrow\frac{y^2}{9}=4\Rightarrow\frac{y}{3}=4\Rightarrow y=4.3=12\)

\(\frac{z^2}{25}=4\Leftrightarrow\frac{z}{5}=4\Rightarrow z=4.5=20\)

vậy x = 8 , y= 12 ,z=20

Answer 2

a) \(\sqrt{3-x}\)=5

=>(\(\sqrt{3-x}\))²=5²

=>3-x=25

=>x=-22

Answer 3

b)\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{81}{64}\)

=> \(\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{9}{8}\right)^2\)

=> \(x+\dfrac{1}{2}=\dfrac{9}{8}\)

=> \(x=\dfrac{9}{8}-\dfrac{1}{2}\)

=> \(x=\dfrac{5}{8}\)

c) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(x^2-2y^2+z^2=44\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

=> \(\left[{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có:

\(x^2-2y^2+z^2=44\)

=> \(\left(2k\right)^2-2\left(3k\right)^2+\left(5k\right)^2=44\)

=> \(4k^2-2.9k^2+25k^2=44\)

=> \(4k^2-18k^2+25k^2=44\)

=> \(\left(4-18+25\right).k^2=44\)

=> \(11.k^2=44\)

=> \(k^2=4\) => \(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

*Với k = 2, ta có:

_\(x=2.2=4\)

_\(y=3.2=6\)

_\(z=5.2=10\)

*Với k = -2, ta có:

_\(x=2.\left(-2\right)=-4\)

_\(y=3.\left(-2\right)=-6\)

_\(z=5.\left(-2\right)=-10\)