(1/3x + 2y) x (1/9x2 - 2/3xy + 4y2)
1,phân tích mỗi đa thức sau thành phân tử
a,(x+2y)2-(x-y)2
b,(x+1)3+(x-1)3
c,9x2-3x+2y-4y2
d,4x2-4xy+2x-y+y2
e,x3+3x2+3x+1-y3
g,x3-2x2y+xy2-4x
a) \(\left(x+2y\right)^2-\left(x-y\right)^2=\left(x+2y+x-y\right)\left(x+2y-x+y\right)\)
\(=\left(2x+y\right).3y\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3\)
\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)
\(=2x\left[\left(x+1\right)^2-\left(x^2-1\right)+\left(x-1\right)^2\right]\)
c) \(9x^2-3x+2y-4y^2\)
\(=9x^2-4y^2-3x+2y\)
\(=\left(3x-2y\right)\left(3x+2y\right)-\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left[3x+2y-1\right]\)
d) \(4x^2-4xy+2x-y+y^2\)
\(=4x^2-4xy+y^2+2x-y\)
\(=\left(2x-y\right)^2+2x-y\)
\(=\left(2x-y\right)\left(2x-y+1\right)\)
e) \(x^3+3x^2+3x+1-y^3\)
\(=\left(x+1\right)^3-y^3\)
\(=\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]\)
g) \(x^3-2x^2y+xy^2-4x\)
\(=x\left(x^2-2xy+y^2\right)-4x\)
\(=x\left(x-y\right)^2-4x\)
\(=x\left[\left(x-y\right)^2-4\right]\)
\(=x\left(x-y+2\right)\left(x-y-2\right)\)
a) (x + 2y)² - (x - y)²
= (x + 2y - x + y)(x + 2y + x - y)
= 3y(2x + y)
b) (x + 1)³ + (x - 1)³
= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]
= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)
= 2x(x² + 3)
c) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) x³ + 3x² + 3x + 1 - y³
= (x³ + 3x² + 3x + 1) - y³
= (x + 1)³ - y³
= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]
= (x - y + 1)(x² + 2x + 1 + xy + y + y²)
g) x³ - 2x²y + xy² - 4x
= x(x² - 2xy + y² - 4)
= x[(x² - 2xy + y²) - 4]
= x[(x - y)² - 2²]
= x(x - y - 2)(x - y + 2)
Cho hệ 9 x 2 − 4 y 2 = 5 log m 3 x + 2 y − log 3 3 x − 2 y = 1 có nghiệm x ; y thỏa mãn 3 x + 2 y ≤ 5. Khi đó giá trị lớn nhất của m là
A. -5
B. log 3 5
C. 5
D. log 5 3
Đáp án C
Ta có: 9 x 2 − 4 y 2 = 5 ⇔ 3 x + 2 y 3 x − 2 y = 5 ⇔ 3 x − 2 y = 5 3 x + 2 y
Khi đó: log m 3 x + 2 y = log 3 3 x − 2 y = 1
⇔ log m 3 x + 2 y − log 3 5 3 x + 2 y = 1
⇔ log m 3 x + 2 y + log 3 3 x + 2 y − log 3 5 = 1 ⇔ log m 3. log 3 3 x + 2 y + log 3 3 x + 2 y = log 3 15 ⇔ log 3 3 x + 2 y 1 + log m 3 = log 3 15
Vì 3 x + 2 y ≤ 5
nên log 3 3 x + 2 y ≤ log 3 5 ⇒ log 3 15 1 + log m 3 ≤ log 3 5
⇔ log 3 15 log 3 5 ≤ 1 + log m 3
⇔ log m 3 ≥ log 5 15 − 1 = log 5 3 ⇔ m ≤ 5.
Bài 2: Rút gọn biểu thức
A=(x-2)(x2+2x+4)-(128+x3)
B=(2x+3y)(4x2-6xy+9y2)-(3x-2y)(9x2+6xy+4y2)
\(A=x^3-8-128-x^3=-136\\ B=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)
\(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(128+x^3\right)=x^3-8-128-x^3=-136\)
\(B=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)-\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)
\(A=x^3+2x^2+4x-2x^2-4x-8-128-x^3\)
\(A=-136\)
\(B=\left(2x+3y\right)\left(2x-3y\right)^2-\left(3x-2y\right)\left(3x+2y\right)^2\)
\(B=\left(2x+3y\right)\left(2x-3y\right)\left(2x-3y\right)-\left(3x-2y\right)\left(3x+2y\right)\left(3x+2y\right)\)
\(B=\left(4x^2-9y^2\right)\left(2x-3y\right)-\left(9x^2-4y^2\right)\left(3x+2y\right)\)
\(B=8x^3-12x^2y-18xy^2-27y^3-27x^3-18x^2y+12xy^2+8y^3\)
\(B=-19x^3-30x^2y-6xy^2-19y^3\)
\(B=-19\left(x^3-y^3\right)-6xy\left(5x+y\right)\)
Tính giá trị của phân thức A = 3 x - 2 y 3 x + 2 y biết rằng 9 x 2 + 4 y 2 = 20 x y , và 2y < 3x < 0.
phân tích đa thức thành nhân tử 2 ẩn :
a) 2x2+xy-y2-x+2y-1
b) 3x2-2xy-y2-10x-2y+3
c) 3x2y-xy2+xy-2y2-3x-9y+5
d) 2x2y2-3xy-2y2+y+1
e) 3x3-12xy2-5x2-4y2+x+1
a)2x^2+xy-y^2-x+2y-1
=2x^2+xy-x-(y-1)^2
=2x^2+x(y-1)-(y-1)^2
=2a^2+ab-b^2 với a=x,b=y-1
=2a^2+2ab-ab-b^2
=(2a-b)(a+b)
=(2x-y+1)(x+y-1)
Tính giá trị của phân thức A=\(\dfrac{3x-5y}{3x+7y}\), biết rằng 9x2+4y2=20xy, và 2y<3x<0
Tính giá trị của biểu thức:
a) N= (25x2 + 10xy + 4y2)(5x - 2y) tại x=1/5:y=1/2
b) Q= (x + 3y)(x2 - 3xy + 9y2) tại x=y=1/2
Giúp mik với ạ!
a: \(N=\left(5x\right)^3-\left(2y\right)^3=1^3-1^3=0\)
b: \(Q=x^3+27y^3=\dfrac{1}{8}+\dfrac{27}{8}=\dfrac{28}{8}=\dfrac{7}{2}\)
1) Rut gon cac bieu thuc sau :
a) (x-3)(x2+3x+ 9)-(54+x3)
b) (3x+y)(9x2-3xy +y2)-(3x-y)(9x2+3xy+y2)
2, Dien cac don thuc thich hop vao cho trong
a, (x+3y) (... - ... + ...) = x^3 +27y^3
b, (2x- ....) (... + 6xy + ... +...) = 8x^3 - 27y^3
\(1.\)
\(a.\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)
\(=\left(x^3-3^3\right)-\left(54+x^3\right)\)
\(=x^3-27-54-x^3\)
\(=-81\)
\(b.\)
\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left(27x^3+y^3\right)-\left(27x^3-y^3\right)\)
\(=27x^3+y^3-27x^3+y^3\)
\(=2y^3\)
\(2.\)
\(a.\)
\(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
\(b.\)
\(\left(2x-3y\right)\left(4x^2+6xy+9y^3\right)=8x^3-27y^3\)
1) a) \(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)
\(=\left(x^3-3^3\right)-\left(54+x^3\right)\\ =\left(x^3-27\right)-54-x^3\\ =-27-54\\ =-81\)
b) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left[\left(3x\right)^3+y^3\right]-\left[\left(3x\right)^3-y^3\right]\\ =2y^3\)
2) a) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
b) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=8x^3-27y^3\)
Làm phép tính
a)x/x-2y+x/x+2y+4xy/4y2-x2
b)4x+7/2x+2-3x+6/2x+2
c)x+9/x2-9-3/x2+3x
d)1/x2+3x+2-1/x2-4
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
Giúp mik vs cần gấp!!!
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
\(d,\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
\(\Rightarrow\)PT vô nghiệm vì 11 không phải là tổng 2 số chính phương