Câu 1: D

Câu 2: B

Câu 3: D

Câu 4: C

Câu 5: A

Lời giải:

Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$

$\Rightarrow x=at; y=bt; z=ct$. Ta có:

$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$

Mặt khác:

$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$

Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)

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Câu hỏi của Cao Thành Lộc - Toán lớp 7 - Học toán với OnlineMath

Giúp mk nhanh nha , mai mk phải kiểm tra rùi

Ta có: \(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Leftrightarrow3x-y+2x+y=10z\)

\(\Leftrightarrow5x=10z\)

hay x=2z

Thay x=2z vào biểu thức 3x-y=3z, ta được:

\(3\cdot2z-y=3z\)

\(\Leftrightarrow6z-y=3z\)

hay y=3z

Thay x=2z và y=3z vào biểu thức \(M=\dfrac{x^2-2xy}{x^2+y^2}\), ta được:

\(M=\dfrac{\left(2z\right)^2-2\cdot2z\cdot3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{13z^2}=\dfrac{-8z^2}{13z^2}=\dfrac{-8}{13}\)

Vậy: \(M=\dfrac{-8}{13}\)

\(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5x=10z\\3x-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\3.2z-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=3.2z-3z=6z-3z=3z\end{matrix}\right.\)

Có: \(M=\dfrac{x^2-2xy}{x^2+y^2}=\dfrac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{4z^2+9z^2}=\dfrac{-8z^2}{13z^2}==-\dfrac{8}{13}\)

\(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5x=10z\\3x-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\3.2z-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=3.2z-3z=6z-3z=3z\end{matrix}\right.\)

Có: \(M=\dfrac{x^2-2xy}{x^2+y^2}=\dfrac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{4z^2+9z^2}=\dfrac{-8z^2}{13z^2}==-\dfrac{8}{13}\)

Lời giải:

PT $\Leftrightarrow \frac{a+b-x}{c}+1+\frac{a+c-x}{b}+1+\frac{b+c-x}{a}+1+\frac{4x}{a+b+c}-4=0$

$\Leftrightarrow \frac{a+b+c-x}{c}+\frac{a+b+c-x}{b}+\frac{a+b+c-x}{a}-\frac{4(a+b+c-x)}{a+b+c}=0$

$\Leftrightarrow (a+b+c-x)(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c})=0$

$\Rightarrow a+b+c-x=0$ hoặc $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$
Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$, khi đó $x$ nhận mọi giá trị thực.

Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}\neq 0$

$\Rightarrow a+b+c-x=0$
$\Rightarrow x=a+b+c$

\(VT=\dfrac{1}{\left(x-y\right)^2}+\dfrac{x^2+y^2}{x^2y^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x-y\right)^2+2xy}{x^2y^2}\)

\(VT=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x-y\right)^2}{x^2y^2}+\dfrac{2}{xy}\ge2\sqrt{\dfrac{\left(x-y\right)^2}{\left(x-y\right)^2x^2y^2}}+\dfrac{2}{xy}=\dfrac{2}{\left|xy\right|}+\dfrac{2}{xy}\ge\dfrac{2}{xy}+\dfrac{2}{xy}=\dfrac{4}{xy}\)

a) \(ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c}\)

\(ab=c^2\Rightarrow\frac{a}{c}=\frac{c}{b}\)

Suy ra: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)

\(P=\frac{b^{333}}{a^{111}.c^{222}}=\frac{b^{333}}{a^{111}.c^{111}.c^{111}}=\frac{b^{333}}{\left(ac\right)^{111}.c^{111}}=\frac{b^{333}}{\left(b^2\right)^{111}.c^{111}}=\frac{b^{333}}{b^{222}.c^{111}}=\frac{b^{111}}{c^{111}}=\left(\frac{b}{c}\right)^{111}\)

\(=1^{111}=1\)

Giúp mình vớiiiiiiiiiiiiii

Điều kiện xác định là `{(x-3 ne 0),(x(x-3) ne 0):}`

                 `<=>{(x ne 3),(x ne 0):}`

      `=>bb A`

ĐCXĐ: \(\left\{{}\begin{matrix}x\ne0\\x-3\ne0\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)