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2x.(x - 5) - x.(3 + 2x) = 26
=> (2x2 - 10x) - (3x + 2x2) = 26
=> 2x2 - 10x - 3x - 2x2 = 26
=> -13x = 26
=> x = 26 : (-13)
=> x = -2
2x.﴾x ‐ 5﴿ ‐ x.﴾3 + 2x﴿ = 26 => ﴾2x 2 ‐ 10x﴿ ‐ ﴾3x + 2x 2 ﴿ = 26 => 2x 2 ‐ 10x ‐ 3x ‐ 2x 2 = 26 => ‐13x = 26 => x = 26 : ﴾‐13﴿ => x = ‐2
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Ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2a+2b+2c}{a+b+c}=2\)
\(\Rightarrow\) a + b = 2c; b + c = 2a; c + a = 2b
\(\Rightarrow\) M = \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
= \(\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{a+c}{a}\right)\)
= \(\frac{2c}{b}\times\frac{2a}{c}\times\frac{2b}{a}\)
= 8
Vậy: M = 8.
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\(\left\{{}\begin{matrix}x-y-z=0\\x+2y-10z=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=3z\\x=y+z=4z\\x+2y=10z\end{matrix}\right.\)
\(B=\dfrac{2x^2+4xy}{y^2+z^2}=\dfrac{2x\left(x+2y\right)}{9z^2+z^2}=\dfrac{2.4z.10z}{10.z^2}=8\)
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điều kiện xác định:
\(x\ne3;x\ne-3\)\(\dfrac{13-x}{x+3}+\dfrac{6x^2+6}{x^4-8x^2-9}-\dfrac{3x+6}{x^2+5x+6}-\dfrac{2}{x-3}=0\\ \Leftrightarrow\dfrac{13-x}{x+3}+\dfrac{6\left(x^2+1\right)}{\left(x^2-9\right)\left(x^2+1\right)}-\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\dfrac{2}{x-3}=0\\\Leftrightarrow\dfrac{13-x}{x+3}+\dfrac{6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x+3}-\dfrac{2}{x-3}=0\\ \Leftrightarrow\dfrac{\left(13-x\right)\left(x-3\right)+6-3\left(x-3\right)-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow16x-39-x^2+6-3x^{ }+9-2x-6=0\\ \Leftrightarrow-x^2-11x-30=0\\ \Leftrightarrow^{ }-\left(x^2+11x+30\right)=0\\ \Leftrightarrow-\left(x+5\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(tmdkxd\right)\\x=-6\left(tmdkxd\right)\end{matrix}\right.\)
Vậy phương trinh có tập nghiệm là S={-5;-6}
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\(\dfrac{x-3}{x-2}=\dfrac{x-2}{x-4}\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=\left(x-2\right)\left(x-2\right)\)
\(\Leftrightarrow x^2-7x+12=x^2-4x+4\)
\(\Leftrightarrow x^2-7x-x^2+4x=-12+4\)
\(\Leftrightarrow-3x=-8\)
\(\Leftrightarrow x=\dfrac{-8}{-3}=\dfrac{8}{3}\)(Oh....La...La Ngày 8/3)
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\(\left\{\begin{matrix}x-y-z=0\left(1\right)\\x+2y-10z=0\left(2\right)\end{matrix}\right.\)
Lấy (1) - (2), ta có:
\(-3y+9z=0\Leftrightarrow-3\left(y-z\right)=0\)
\(\Rightarrow y-z=0\)
\(\Rightarrow y=-z\)
Thay y=-z vào (1), ta có:
\(x-\left(-z\right)-z=0\Rightarrow x=0\)
Thay x=0 vào B, ta được B=0 (tử bằng 0)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài không khó hi vọng bạn có thể đặt bút xuống làm hoặc bấm = máy Casio cũng được
Mình xin chúc các bạn thi giải toán violympic làm bài tốt , 300/300 nhé !