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Đơn giản thôi!!
Từ giả thiết, suy ra
\(\frac{x}{a+2b+c}=\frac{2y}{4a+2b-2c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{9a}\) (1)
\(\frac{2x}{2a+4b+2c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2x+y-z}{9b}\) (2)
\(\frac{4x}{4a+8b+4x}=\frac{4y}{8a+4b-4c}=\frac{z}{4a-4b+c}=\frac{4x-4y+x}{9c}\) (3)
Từ (1) , (2) và (3) suy ra:
\(\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)
\(\frac{9a}{x+2y+z}-\frac{9b}{2x+y-z}=\frac{9c}{4x-4y+z}\)
\(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}^{\left(đpcm\right)}\)
\(a)4a\left(b-c+2a\right)\)
\(=4ab-4ac+8a^2\)
\(b)-\left(m-n\right)-\left(2m+n-p\right)\)
\(=-m+n-2m-n+p\)
\(=\left(-m-2m\right)+\left(n-n\right)+p\)
\(=p-3m\)
\(c)-\left(x-y\right)+\left(-3x-2y+z\right)\)
\(=-x+y-3x-2y+z\)
\(=\left(-x-3x\right)+\left(y-2y\right)+z\)
\(=z-4x-y\)
\(d)-\left(2a-2b\right)+\left(2a-3b+c\right)\)
\(=-2a+2b+2a-3b+c\)
\(=\left(-2a+2a\right)+\left(2b-3b\right)+c\)
\(=c-b\)
\(4x-xy+2y=3\)
\(\Rightarrow x\left(4-y\right)-8+2y=3-8\)
\(\Rightarrow x\left(4-y\right)-2\left(4-y\right)=-5\)
\(\Rightarrow\left(x-2\right)\left(4-y\right)=-5\)
\(\Rightarrow\left(x-2\right)\left(y-4\right)=5\)
\(\Rightarrow\left(x-2\right);\left(y-4\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Tự xét bảng
\(3y-xy-2x-5=0\)
\(\Rightarrow y\left(3-x\right)-2x=5\)
\(\Rightarrow y\left(3-x\right)+6-2x=5+6\)
\(\Rightarrow y\left(3-x\right)+2\left(3-x\right)=11\)
\(\Rightarrow\left(y+1\right)\left(3-x\right)=11\)
\(\Rightarrow\left(3-x\right);\left(y+1\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
Tự xét
\(2xy-x-y=100\)
\(\Rightarrow x\left(2y-1\right)-y=100\)
\(2x\left(2y-1\right)-\left(2y-1\right)=100+1\)
\(\left(2x-1\right)\left(2y-1\right)=101\)
\(\Rightarrow\left(2x-1\right);\left(2y-1\right)\inƯ\left(101\right)=\left\{\pm1;\pm101\right\}\)
Tự xét bảng
P/s : bài 3 có gì sai ko ?
a) \(\left(x-y\right)-\left(x-z\right)=\left(z+x\right)-\left(y+x\right)\)
BL:
Ta có: \(\left(x-y\right)-\left(x-z\right)\)
\(=x-y-x+z\)
\(=z+x-y-x\)
\(=\left(z+x\right)-\left(y+x\right)\)
\(\Rightarrow\) \(\left(x-y\right)-\left(x-z\right)=\left(z+x\right)-\left(y+x\right)\)
b) \(\left(x-y+z\right)-\left(y+z-x\right)-\left(x-y\right)=\left(z-y\right)-\left(z-x\right)\)
BL:
Lại có: \(\left(x-y+z\right)-\left(y+z-x\right)-\left(x-y\right)\)
\(=x-y+z-y-z+x-x+y\)
\(=\left(x-y-x+y\right)+\left(z-y\right)-\left(z-x\right)\)
\(=\left(z-y\right)-\left(z-x\right)\)
\(\Rightarrow\) \(\left(x-y+z\right)-\left(y+z-x\right)-\left(x-y\right)=\left(z-y\right)-\left(z-x\right)\)
c) \(a\left(b+c\right)-b\left(a-c\right)=\left(a+b\right)c\) BL: Ta lại có: \(a\left(b+c\right)-b\left(a-c\right)=\left(a+b\right)c\) \(=ab+ac-ba+bc\) \(=\left(ab-ba\right)+\left(ac+bc\right)\) \(=0+\left(a+b\right)c\) \(=\left(a+b\right)c\) \(\Rightarrow\) \(a\left(b+c\right)-b\left(a-c\right)=\left(a+b\right)c\) \(\rightarrow\) đpcm.
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