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a . \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-4^2\)
b . \(\left(x-y+6\right)\left(x+y-6\right)=x^2-\left(y-6\right)^2\)
c . \(\left(y+2z-3\right)\left(y-2z-3\right)=\left(y-3\right)^2-\left(2z\right)^2\)
d . \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}\)
\(=\frac{\left(x^2-36\right).3}{\left(2x+10\right)\left(6-x\right)}\)
\(=\frac{3\left(x+6\right)\left(x-6\right)}{\left(2x+10\right)\left(6-x\right)}\)
\(=-\frac{3\left(x+6\right)\left(x-6\right)}{2\left(x+5\right)\left(x-6\right)}\)
\(=-\frac{3\left(x+6\right)}{2\left(x+5\right)}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
a, \(A=x\left(x+y\right)-x\left(y-x\right)=x^2+xy-xy+x^2=2x^2\)
Thay x vào ta có : \(2\left(-3\right)^2=2.9=18\)
y bị lược bỏ rồi mà bạn hay chỗ x^2 + xy - xy + x^2 thay vào à ? lạ !?!
b, \(B=4x\left(2x+y\right)+2y\left(2x+y\right)-y\left(y+2x\right)=8x^2+4xy+4xy+2y^2-y^2-2xy\)
\(=8x^2+6xy+y^2\)
Thay x = 1/2 ; y = -3/4 ta có : Tự thay nhé -> P/s
a)
\(A=x.\left(x+y\right)-x.\left(y-x\right)\)
\(A=x^2+x.y-x.y+x^2\)
\(A=2.x^2\)
Thay x= -3 vào biểu thức A ta được ;
\(A=2.\left(-3\right)^2=2.9=18\)
b) \(B=4.x\left(2x+y\right)+2y\left(2x+y\right)-y\left(y+2x\right)\)
\(B=4x\left(2x+y\right)+2y\left(2x+y\right)-y\left(2x+y\right)\)
\(B=\left(2x+y\right).\left(4x+2y-y\right)\)
\(B=\left(2x+y\right).\left(4x+y\right)\)
\(B=8x^2+2xy+4xy+y^2\)
\(B=8x^2+6xy+y^2\)
Thay \(x=\frac{1}{2}\) và \(y=\frac{-3}{4}\) vào biểu thức B ta được :
\(B=8.\left(\frac{1}{2}\right)^2+6.\frac{1}{2}.\left(\frac{-3}{4}\right)+\left(\frac{-3}{4}\right)^2\)
\(B=2+\left(\frac{-9}{4}\right)+\frac{9}{16}=\frac{5}{16}\)
Bài 2 :
\(A=4\left(x-6\right)-5x\left(x+1\right)+8\left(x^2-x-2\right)\)
\(A=4x-24-5x^2-5x+8x^2-8x-16\)
\(A=-9x-40+3x^2\)
Thay x=-1 vào biểu thức A ta được :
\(A=-9.\left(-1\right)-40+3.\left(-1\right)^2\)
\(A=9-40+3=-28\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^
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a, \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-4^2\)
b, \(\left(y+2z-3\right)\left(y-2z-3\right)=\left(y-3+2z\right)\left(y-3-2z\right)=\left(y-3\right)^2-\left(2z\right)^2\)
c, \(\left(x-y-6\right)\left(x+y-6\right)=\left(x-6-y\right)\left(x-6+y\right)=\left(x-6\right)^2-y^2\)
d, \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z+x\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)