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Bài 2:
a: \(=xy+xz-xy+yz\)
\(=xz+yz=z\left(x+y\right)\)
b: \(=x\left(y-z-y-a\right)\)
\(=x\left(-z-a\right)=-x\left(z+a\right)\)
Bài 1:
a) Chỗ y6 là 6.y hay là y6
b) \(2\left(x-1\right)-3\left(2x+2\right)-4\left(2x+3\right)=16\)
\(\Rightarrow2x-2-6x-6-8x-12=16\)
\(\Rightarrow\left(2x-6x-8x\right)-\left(2+6+12\right)=16\)
\(\Rightarrow-12x-20=16\)
\(\Rightarrow-12x=36\)
\(\Rightarrow x=-3\)
Vậy x = -3
c) \(\left(x-5\right)^{x+1}-\left(x-5\right)^{x+13}=0\)
\(\Rightarrow\left(x-5\right)^{x+1}\left[1-\left(x-5\right)^{12}\right]=0\)
\(\Rightarrow\left(x-5\right)^{x+1}=0\) hoặc \(1-\left(x-5\right)^{12}=0\)
+) \(\left(x-5\right)^{x+1}=0\Rightarrow x-5=0\Rightarrow x=5\)
+) \(1-\left(x-5\right)^{12}=0\Rightarrow\left(x-5\right)^{12}=1\)
\(\Rightarrow x-5=\pm1\)
+) \(x-5=1\Rightarrow x=6\)
+) \(x-5=-1\Rightarrow x=4\)
Vậy \(x\in\left\{6;4\right\}\)
Bài 2: a, thiếu dữ liệu
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)
\(\left[\begin{matrix}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{a}=1\end{matrix}\right.\Rightarrow\left[\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)
Ta có: \(\frac{a^3b^2c^{1930}}{a^{1935}}=\frac{a^3a^2a^{1930}}{a^{1935}}=\frac{a^{1935}}{a^{1935}}=1\)
Vậy \(\frac{a^3b^2c^{1930}}{a^{1935}}=1\)
Vì : \(\left|x+2017\right|\ge0\forall x\)
\(\left|y-2017\right|\ge0\forall y\)
\(\Rightarrow\left|x+2017\right|+\left|y-2017\right|\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+2017=0\\y-2017=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2017\\y=2017\end{matrix}\right.\)
Vậy x = -2017 ; y = 2017
b, \(\left|2x-1\right|=\left|x+8\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=x+8\\2x-1=-\left(x+8\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x-x=8+1\\2x+x=-8+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\3x=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{-7}{3}\end{matrix}\right.\)
Vậy x = 9
c, \(\left|3x-2\right|-\left|x+14\right|=0\Rightarrow\left|3x-2\right|=\left|x+14\right|\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=x+14\\3x-2=-\left(x+14\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=14+2\\3x+x=-14+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=16\\4x=-12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
Vậy ...
Làm 1 câu thôi nha bạn,mỏi tay lắm:
\(\left|x+2017\right|+\left|y-2017\right|=0\)
\(\left|x+2017\right|\ge0\)
\(\left|x-2017\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left|x+2017\right|=0\Rightarrow x+2017=0\Rightarrow x=-2017\)
\(\left|y-2017\right|=0\Rightarrow y-2017=0\Rightarrow y=2017\)
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