1 phân tích đa thức thành nhân tử
a,\(3x^2-6xy+3y^2\)
b,\(\left(x-y\right)^2-4x^2\)
2.tìm x biết
a,2x(x-3)-x+3=0
b,\(x^2+5x+6=0\)
Bài 1: Phân tích đa thức thành nhân tử:
1) \(3x^3y^2-6xy\)
2) \(\left(x-2y\right).\left(x+3y\right)-2.\left(x-2y\right)\)
3) \(\left(3x-1\right).\left(x-2y\right)-5x.\left(2y-x\right)\)
4) \(x^2-y^2-6y-9\)
5) \(\left(3x-y\right)^2-4y^2\)
6) \(4x^2-9y^2-4x+1\)
8) \(x^2y-xy^2-2x+2y\)
9) \(x^2-y^2-2x+2y\)
Bài 2: Tìm x:
1) \(\left(2x-1\right)^2-4.\left(2x-1\right)=0\)
2) \(9x^3-x=0\)
3) \(\left(3-2x\right)^2-2.\left(2x-3\right)=0\)
4) \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
Bài 1 : Phân tích đa thức thành nhân tử
a) 5x^2y-20xy^2
b) 1-8x+16x^2-y^2
c) 4x-4-x^2
d) x^3-2x^2+x-xy^2
e)27-3x^2
f) 2x^2+4x+2-2y^2
Bài 2: tìm x, biết
a) x^2(x-2023)-2023+x=0
b) -x(x-4)+(2x^3-4x^2-9x):x=0
c) x^2+2x-3x-6=0
d) 3x(x-10)-2x+20=0
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
Bài 2
a) x²(x - 2023) - 2023 + x = 0
x²(x - 2023) - (x - 2023) = 0
(x - 2023)(x² - 1) = 0
x - 2023 = 0 hoặc x² - 1 = 0
*) x - 2023 = 0
x = 2023
*) x² - 1 = 0
x² = 1
x = 1 hoặc x = -1
Vậy x = -1; x = 1; x = 2023
b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0
-x² + 4x + 2x² - 4x - 9 = 0
x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
Vậy x = 3; x = -3
c) x² + 2x - 3x - 6 = 0
(x² + 2x) - (3x + 6) = 0
x(x + 2) - 3(x + 2) = 0
(x + 2)(x - 3) = 0
x + 2 = 0 hoặc x - 3 = 0
*) x + 2 = 0
x = -2
*) x - 3 = 0
x = 3
Vậy x = -2; x = 3
d) 3x(x - 10) - 2x + 20 = 0
3x(x - 10) - (2x - 20) = 0
3x(x - 10) - 2(x - 10) = 0
(x - 10)(3x - 2) = 0
x - 10 = 0 hoặc 3x - 2 = 0
*) x - 10 = 0
x = 10
*) 3x - 2 = 0
3x = 2
x = 2/3
Vậy x = 2/3; x = 10
bài 1: phân tích đa thức thành nhân tử
a,2x+10y
b,x\(^2+4x+4\)
c,\(x^2-y^2+10y-25\)
bài 2 tìm x, biết
a,\(x^2-3x+x-3=0\)
b,\(2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\)
c,\(x^2-\left(x-3\right)\left(2x-5\right)=9\)
\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)
\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
Bài 1. Phân tích các đa thức sau thành nhân tử a) y - 9 - x + 6x b) 25 - 4x? - 4xy - y c) x - xz + 4y - 2yz + 4xy d) 3x + 6xy - 48z + 3y? e) x - z + 4y - 4t - 4xy + 4zt f) +2x'y+xy-16x Bài 2. Tìm x biết a) 3x(-3)-4x+12 -0 b) -5x=0 c) (a-2 -(x+2 =0 d) -9-4x+3)=0 Bài 3. Tính nhanh giá trị biểu thức a) A= x - 4z? - 2xy + y với x = -16; y = -6; z = 45 b) B = x - y + 2y-1 với x = 75; y = 26. c) C = 2x + xy - x'y - 2y với x= y =
giúp e làm vs ạ em đang cần gấp
bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được
Phân tích mỗi đa thức sau thành nhân tử
a)x^3-2x^2y+xy^2+xy
b)x^3+4x^2y+4xy^2-9x
c)x^3-y^3+x-y
d)4x^2-4xy+2x-y+y^2
e)9x^2-3x+2y-4y^2
f)3x^2-6xy+3y^2-5x+5y
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
Phân tích đa thức thành nhân tử
\(a.\left(x^2+4x-3\right)^2-5x\left(x^2+4x-3\right)+6x^2\)
B. \(x^2-2xy+y^2+3x-3y-4\)
\(c.\left(12x^2-12xy+3y^2\right)-10\left(2x-y\right)+8\)
\(d.\left(x^2-2x\right)\left(x^2-2x-1\right)-6\)
Phân tích các biểu thức sau thành tích:
a) \(y^2\left(x^2+y\right)-x^2z-yz\)
b) \(\left(2x^2+1\right)\left(3x-2\right)+\left(x-2\right)\left(2-3x\right)+2-3x\)
c) \(\left(x^2-x+2\right)\left(x-1\right)-x^2\left(1-x\right)^2-\left(2x+1\right)\left(1-x\right)^3\)
Tìm x thỏa mãn điều kiện:
a) \(5x^2\left(2x-3\right)+\left(2x^2+3x+3\right)\left(3-2x\right)=6x^3-9x^2\)
b) \(\left(4x^2+2x\right)\left(x^2-x\right)+\left(4x^2+6\right)\left(x-x^2\right)=0\)
c) Phân tích đa thức: \(x^{m+3}y^2-3x^3y^{m+5}\)thành nhân tử
3A. Tính giá trị biểu thức: a) A = (x²-3x² + 3x)² -2(x²-3x² + 3x)+1 tại x= 11; b) B=(x-2y)(x² + 2xy + 4y²)-6xy(x-2y) tai x=3;y=; 5A. Phân tích đa thức thành nhân tử a) x² +1-2x²; c) y²-4x² + 4x-1; b)x²-y²-5y+5x; d) x (2+x)²-(x+2)+1-x² 6A. Phân tích đa thức thành nhân tử: (a) x² −8x+7; b) 2x² -5x+2; c) x²-5x² +8x-4; d) x² +64.
1) Tìm x a. (4x - 1 )² - 8 ( x - 2 ) ( 3 + 2x ) = 15 b. ( 5x - 2 ) ( x - 2 ) - 4 ( x - 3 ) = x² + 3 2) pt đa thức thành nhân tử a. 5x² - 10xy + 5y² - 20z² b. a³ - ay - a²x + xy c. 3x² - 6xy + 3y² - 12z² d. x² - y² + 2yz - z² e. x² - 2xy + tx - 2ty mình cảm ơn ạ
1)
a) => 16x2 - 8x + 1 - 8(2x2 + 3x - 4x - 6) = 15
=> 16x2 - 8x + 1 - 8(2x2 - x - 6) = 15
=> 16x2 - 8x + 1 - 16x2 + 8x + 48 = 15
=> 49 = 15 (?) (vô lí)
=> Không tìm được x thoả mãn
b) (5x - 2)(x - 2) - 4(x - 3) = x2 + 3
=> 5x2 - 10x - 2x + 4 - 4x + 12 = x2 + 3
=> 5x2 - 16x + 16 = x2 + 3
=> 4x2 - 16x + 16 = 3
=> (2x)2 - 2.2x.4 + 42 = 3
=> (2x - 4)2 = 3
=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)
Mong bạn xem lại đề bài!
2)
a) 5x2 - 10xy + 5y2 - 20z2
= 5(x2 - 2xy + y2 - 4z2)
= 5[(x - y)2 - (2z)2]
= 5(x - y - 2z)(x - y + 2z)
b) a3 - ay - a2x + xy
= a(a2 - y) - x(a2 - y)
= (a - x)(a2 - y)
c) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= 3[(x - y)2 - (2z)2]
= 3(x - y - 2z)(x - y + 2z)
d) x2 - 2xy + tx - 2ty
= x(x - 2y) + t(x - 2y)
= (x + t)(x - 2y)