1 a) 2x = 5y và x2+y2=133
giúp mik với T.T
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1: \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
2: \(x^2-y^2+x-y\)
\(=\left(x^2-y^2\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+1\right)\)
3: \(3x-3y+x^2-y^2\)
\(=\left(3x-3y\right)+\left(x^2-y^2\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
4: \(5x-5y+x^2-y^2\)
\(=\left(5x-5y\right)+\left(x^2-y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(5+x+y\right)\)
5: \(x^2-5x-y^2-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
6: \(x^2-y^2+2x-2y\)
\(=\left(x^2-y^2\right)+\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+2\right)\)
7: \(x^2-4y^2+x+2y\)
\(=\left(x^2-4y^2\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+1\right)\)
8: \(x^2-y^2-2x-2y\)
\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
9: \(x^2-4y^2+2x+4y\)
\(=\left(x^2-4y^2\right)+\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+2\right)\)
a) \(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)
b) \(=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\)
c) \(=3x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(3x+5\right)\)
d) \(=\left(x+y\right)^2-25=\left(x+y-5\right)\left(x+y+5\right)\)
e) \(=x\left(x^2-11x+30\right)\)
f) \(=x\left(x-3\right)+6\left(x-3\right)=\left(x-3\right)\left(x+6\right)\)
\(a,=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\\ b,=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\\ c,=x\left(x+1\right)+a\left(x+1\right)=\left(x+a\right)\left(x+1\right)\\ d,Sửa:x^2y+xy^2-3x-3y=xy\left(x+y\right)-3\left(x+y\right)=\left(xy-3\right)\left(x+y\right)\\ e,=xy\left(x+1\right)-\left(x+1\right)=\left(xy-1\right)\left(x+1\right)\\ f,=x^2-4=\left(x-2\right)\left(x+2\right)\\ g,=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\\ h,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ i,=\left(x-4\right)^2-24y^2=\left(x-2\sqrt{6}y-4\right)\left(x+2\sqrt{6}y+4\right)\)
a: =x^3+8-1+27x^3=28x^3+7
b: Sửa đề: (2+y)(y^2-2y+4)+(5-y)(25+5y+y^2)
=8+y^3+125-y^3
=133
b: (x-y)(x^2-2x+y)
\(=x^3-2x^2+xy-x^2y+2xy-y^2\)
\(=x^3-2x^2-x^2y+3xy-y^2\)
c: \(\left(x^2-y\right)\left(x+y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y^2-xy-y^3-\left(x^3-y^3\right)\)
\(=x^2y^2-xy\)
d: \(3x\left(2xy-z\right)-5y\left(x^2-2\right)+3xz\)
\(=6x^2y-3xz-5x^2y+10y+3xz\)
\(=x^2y+10y\)
Mình tính thẳng ra nhé.
a) -A+B-C= -4x^2 + 2xy - 3y^2 + 3y + 7.
b) A+B-(-C)= -5y^2 = 2xy - 4x + 9y + 5.
a, \(5x-xy+y^2-5y\)
\(=x\left(5-y\right)-y\left(5-y\right)\)
\(=\left(5-y\right)\left(x-y\right)\)
b, Có: \(x^2+2x+1-y^2\)
<=> \(\left(x+1\right)^2-y^2\)
<=> \(\left(x+1-y\right)\left(x+1+y\right)\)
Với x = 84; y = 15 ta có:
\(\left(x+1-y\right)\left(x+1+y\right)=\left(84+1-15\right)\left(84+1+15\right)\)
<=> \(70.100=7000\)
\(a.2x^3+6x=2x\left(x^2+3\right)\)
\(=2x\left(x^2+3\right)-2x\left(x^2+3\right)\)
\(=\left(x^2+3\right)\left(2x-2x\right)\)
\(b.5x\left(x-2\right)-3x^2\left(x-2\right)\)
\(=\left(x-2\right)\left(5x-3x^2\right)\)
\(c.3x\left(x-5y\right)-2y\left(5y-x\right)\)
\(=3x\left(x-5y\right)+2\left(x-5y\right)\)
\(=\left(x-5y\right)\left(3x+2\right)\)
\(d.y^2\left(x^2+y\right)-x^3-xy\)
\(=y^2\left(x^2+y\right)-x\left(x^2+y\right)\)
\(=\left(x^2+y\right)\left(y^2-x\right)\)
e. Cái bài này ghi lại đàng hoàng xíu nha t k hỉu
\(f.3x^2\left(y^2-2x\right)-15x\left(2x-y^2\right)\)
\(=3x^2\left(y^2-2x\right)+15x\left(y^2-2x\right)\)
\(=\left(y^2-2x\right)\left(3x^2+15x\right)\)
Ta có : 2x = 5y \(\Rightarrow\frac{x}{5}=\frac{y}{2}\Rightarrow\frac{x^2}{25}=\frac{y^2}{4}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có :
\(\frac{x^2}{25}=\frac{y^2}{4}=\frac{x^2+y^2}{25+4}=\frac{133}{29}\)
Suy ra :
+) \(\frac{x^2}{25}=\frac{133}{29}\Leftrightarrow x^2=\frac{3325}{29}\Leftrightarrow x=\sqrt{\frac{3325}{29}}\approx10,7\)
+)\(\frac{y^2}{4}=\frac{133}{29}\Leftrightarrow y^2=\frac{532}{29}\Leftrightarrow y=\sqrt{\frac{532}{29}}\approx4,3\)
Vậy \(x\approx10,7;y\approx4,3\)