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2.
a. Ta có: x + y = 5 ⇒ x = 5 - y
Thay vào A ta được:
\(A=3\left(5-y\right)^2+3y^2-2y+6\left(5-y\right).y-100\)
\(A=75-30y+3y^2+3y^2-2y+30y-6y^2-100\)
\(A=75-100=-25\)
b. Ta có: x - y = 7 ⇒ x = 7 + y
Thay x = 7 + y vào A ta được:
\(A=\left(7+y\right)\left(7+y+2\right)+y\left(y-2\right)-2\left(7+y\right).y+37\)
\(A=y^2+16y+63+y^2-2y-14y-2y^2+37\)
\(A=100\)
c. Ta có: x + 2y = 5 ⇒ x = 5 - 2y
Thay vào A ta có:
\(A=\left(5-2y\right)^2+4y^2-2\left(5-2y\right)+10+4\left(5-2y\right).y-4y\)
\(A=25-20y+4y^2+4y^2-19+4y+10+20y-8y^2-4y\)
\(A=16\)
Bài 1:
a) \(25\left(x+2y\right)^2-16\left(2x-y\right)^2\)
\(=\left[5\left(x+2y\right)\right]^2-\left[4\left(2x-y\right)\right]^2\)
\(=\left[5\left(x+2y\right)-4\left(2x-y\right)\right]\left[5\left(x+2y\right)+4\left(2x-y\right)\right]\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(14y-3x\right)\left(13x+6y\right)\)
b) \(0,25\left(x-2y\right)^2-4\left(x+y\right)^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)-2\left(x+y\right)\right]\left[\dfrac{1}{2}\left(x-2y\right)+2\left(x+y\right)\right]\)
\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)
\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)
\(=-3\left(\dfrac{1}{2}x+y\right)\left(\dfrac{5}{2}x+y\right)\)
c) \(\dfrac{4}{9}\left(x-3y\right)^2-0,04\left(x+y\right)^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)\right]^2-\left[\dfrac{1}{5}\left(x+y\right)\right]^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)-\dfrac{1}{5}\left(x+y\right)\right]\left[\dfrac{2}{3}\left(x-3y\right)+\dfrac{1}{5}\left(x+y\right)\right]\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)
\(=\dfrac{1}{5}\left(\dfrac{7}{3}x-11y\right).\dfrac{1}{5}\left(\dfrac{13}{3}x-9y\right)\)
\(=\dfrac{1}{25}\left(\dfrac{7}{3}x-11y\right)\left(\dfrac{13}{3}x-9y\right)\)
d) \(-25x^2+30x-9\)
\(=-\left(25x^2-30x+9\right)\)
\(=-\left[\left(5x\right)^2-2.5x.3+3^2\right]\)
\(=-\left(5x-3\right)^2\)
Bài 2:
a) \(x^3y^2-x^2y^3-2x+2y\)
\(=x^2y^2\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2y^2-2\right)\)
Thay x = -1 và y = -2 vào ta được
\(=\left[-1-\left(-2\right)\right]\left[\left(-1\right)^2\left(-2\right)^2-2\right]\)
\(=1\left(4-2\right)\)
\(=2\)
b) \(5x^2-3x+3y-5y^2\)
\(=5\left(x^2-y^2\right)-3\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
Thay x = 3 và y = 1 vào ta được
\(=5\left(3-1\right)\left(3+1\right)-3\left(3-1\right)\)
\(=5.2.4-3.2\)
\(=34\)
a)
\(A=x^2y-y+xy^2-x\)
\(A=\left(x^2y-x\right)-\left(y-xy^2\right)\)
\(A=x.\left(xy-1\right)-y.\left(1-xy\right)\)
\(A=x.\left(xy-1\right)+y.\left(xy-1\right)\)
\(A=\left(xy-1\right).\left(x+y\right)\)
Thay \(x=-5\) và \(y=2\) vào biểu thức A, ta được:
\(A=\left[\left(-5\right).2-1\right].\left[\left(-5\right)+2\right]\)
\(A=\left(-11\right).\left(-3\right)\)
\(A=33.\)
Vậy giá trị của biểu thức A tại \(x=-5\) và \(y=2\) là \(33.\)
Chúc bạn học tốt!
a) \(A=3x\left(10x^2-2x+1\right)-6x\left(5x^2-x-2\right)\)
\(=30x^3-6x^2+3x-30x^3+6x^2+12x\)
\(=15x\)
Thay \(x=15\) vào biểu thức A.
Ta có: \(15\cdot15=225\)
Vậy giá trị biểu thức A tại \(x=15\) là 225.
b) \(5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2\)
Thay \(x=-\dfrac{1}{5};y=-\dfrac{1}{2}\) vào biểu thức B.
Ta có: \(5\cdot\left(-\dfrac{1}{5}\right)^2-4\cdot\left(-\dfrac{1}{2}\right)^2=-\dfrac{4}{5}\)
Vậy giá trị biểu thức B tại \(x=-\dfrac{1}{5};y=-\dfrac{1}{2}\) là \(-\dfrac{4}{5}\)
a) ĐKXĐ : \(x+y\ne0\)
\(x^2-2y^2=xy\)
\(x^2-y^2-y^2-xy=0\)
\(\left(x-y\right)\left(x+y\right)-y\left(y+x\right)=0\)
\(\left(x+y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\left(Loai\right)\\x-2y=0\left(Chon\right)\end{matrix}\right.\)
Với x - 2y = 0 ta có x = 2y
Thay x = 2y vào A ta có :
\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
Câu 1:
a: \(=\left(5x+10y\right)^2-\left(8x-4y\right)^2\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(-3x+14y\right)\left(13x+6y\right)\)
b: \(=\left(0.5x-y\right)^2-\left(2x+2y\right)^2\)
\(=\left(0.5x-y-2x-2y\right)\left(0.5x-y+2x+2y\right)\)
\(=\left(-1.5y-3y\right)\left(2.5x+y\right)\)
c: \(=\left(\dfrac{2}{3}x-2y\right)^2-\left(0.2x+0.2y\right)^2\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x+2y+\dfrac{1}{5}y+\dfrac{1}{5}x\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x+\dfrac{11}{5}y\right)\)
d: \(=-\left(5x-3\right)^2\)
\(A=\left(5x-2y\right)\left(5x+2y\right)\)
\(A=\left(5x\right)^2-\left(2y\right)^2\)
\(A=25x^2-4y^2\)
\(A=25.\left(-2\right)^2-4\left(-10\right)^2\)
\(A=25.4-4.100\)
\(A=100-400\)
\(A=300\)
\(B=\left(2x-5\right)\left(4x^2+10x+25\right)\)
\(B=\left(2x\right)^3-5^3\)
\(B=8x^3-125\)
\(B=8.8-125\)
\(B=64-125\)
\(B=-61\)
\(C=\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
\(C=\left(3x\right)^2+\left(2y\right)^2\)
\(C=9x^2+4y^2\)
\(C=9\left(-1\right)^2+4\left(\dfrac{1}{2}\right)^2\)
\(C=9+4.\dfrac{1}{4}\)
\(C=9+1\)
\(C=10\)