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Điểm K ở đâu vậy bạn?

ê bạn có ny chx mik k12 nek

Xét ΔABC có \(\hat{BAC}+\hat{ABC}+\hat{ACB}=180^0\)

=>\(\hat{ABC}+\hat{ACB}=180^0-80^0=100^0\)

=>\(2\left(\hat{IBC}+\hat{ICB}\right)=100^0\)

=>\(\hat{IBC}+\hat{ICB}=50^0\)

Xét ΔIBC có \(\hat{IBC}+\hat{ICB}+\hat{BIC}=180^0\)

=>\(\hat{BIC}=180^0-50^0=130^0\)

\(\Rightarrow\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)+\dfrac{1}{2}\left(2x+1\right)=\dfrac{-13}{3}\)

\(\Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}+x+\dfrac{1}{2}=\dfrac{-13}{3}\)

\(\Rightarrow\dfrac{4}{3}x=\dfrac{-13}{3}\Rightarrow x=\dfrac{-13}{4}\)

\(\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

Vậy \(x\in\left\{\pm7\right\}\)

\(\left|x\right|=0\)

\(\Rightarrow x=0\)

Vậy x = 0

mình nhầm m=\(\frac{c+2b-3a}{c-2b+3a}\)

đặt \(k=\frac{a}{3}=\frac{b}{4}=\frac{c}{5}\Rightarrow a=3k,b=4k,c=5k\)

\(M=\frac{5k+2.4k-3.3k}{5k-2.4k+3.3k}=\frac{4k}{6k}=\frac{2}{3}\)

a/

Theo đề,ta có:

+/ \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)

+/\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)\(\left(2\right)\)

Từ (1) và (2), ta có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)

Do đó:

+/ \(\dfrac{x}{8}=\dfrac{28}{-19}\Rightarrow x=-\dfrac{224}{19}\)

+/\(\dfrac{y}{12}=\dfrac{28}{-19}\Rightarrow y=-\dfrac{336}{19}\)

+/\(\dfrac{z}{15}=\dfrac{28}{-19}\Rightarrow z=-\dfrac{420}{19}\)

Vậy: + \(x=-\dfrac{224}{19}\)

+ \(y=-\dfrac{336}{19}\)

+ \(z=-\dfrac{420}{19}\)

a,$\frac{x}{2}=\frac{y}{3},\frac{y}{4}=\frac{z}{5}$và x-y-z=28

Có \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)

=>\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng tính chất DTSBN có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)=\(\dfrac{x-y-z}{8-12-15}=\dfrac{-28}{19}\)

=> x=\(\dfrac{-224}{19}\)

y=\(\dfrac{-336}{19}\)

z=\(\dfrac{-420}{19}\)

1) \(\left|x\right|=7\)

=> \(\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

Vậy \(x\in\left\{7;-7\right\}.\)

2) \(\left|x\right|=0\)

=> \(x=0\)

Vậy \(x\in\left\{0\right\}.\)

5) \(\left|x\right|-1=\frac{2}{5}\)

=> \(\left|x\right|=\frac{2}{5}+1\)

=> \(\left|x\right|=\frac{7}{5}\)

=> \(\left[{}\begin{matrix}x=\frac{7}{5}\\x=-\frac{7}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{7}{5};-\frac{7}{5}\right\}.\)

8) \(\left|x-17\right|=23\)

=> \(\left[{}\begin{matrix}x-17=23\\x-17=-23\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=23+17\\x=\left(-23\right)+17\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=40\\x=-6\end{matrix}\right.\)

Vậy \(x\in\left\{40;-6\right\}.\)

Mình chỉ làm thế thôi nhé, bạn đăng hơi nhiều mà với cả mấy câu này dễ mà bạn.

Chúc bạn học tốt!

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