Tìm x biết
a, (x-2)2016 + I y2 - 9 I2017 = 0
b, x2 - xy + y = 10 ( x;y \(\in\)Z)
1/ phân tích đa thức thành nhân tử
a)5x – 20y
b)5x.(x – 1) – 3x(x – 1)
c) x.(x+y) – 5x – 5y
2/tính giá trị biểu thức
a) X2 + xy + x tại x = 77 , y = 22
b) X . ( x – y ) + y . ( y – x ) tại x = 53 ,y = 3
3/ tìm x biết
a) X + 5x2 = 0
b) X + 1 = ( x + 1 )2
4 / tính nhanh
a) 97 . 13 + 130 . 0,3
b)86 . 153 – 530 . 8,6
C) 85 .12,7 + 5,3 . 12,7
D)52.143 – 52 . 39 – 8.26
1/
a)5x – 20y=5(x-4y)
b) 5x.(x – 1) – 3x(x – 1)=2x(x-1)
c) x.(x+y) – 5x – 5y=c) x.(x+y) – 5(x+y)=(x-5)(x+y)
2/
a)x2 + xy + x = x(x+y+1)=77.(77+22+1)=77.100=7700
b) x . ( x – y ) + y . ( y – x )=(x-y)(x-y)=(x-y)2=(53-3)2=2500
3/
a) X + 5x2 = 0
⇒x(x+5)=0
⇒hoặc x=0
x+5=0⇒x=-5
b)x + 1 = ( x + 1 )2
⇒(x + 1)-( x + 1 )2 =0
⇒x(x+1)=0
⇒ hoặc x=0
hoặc x+1=0⇒x=-1
4/
a) 97 . 13 + 130 . 0,3 = 97.13+13.10.0,3=97.13+13.3=100.13=1300
b)86 . 153 – 530 . 8,6=86.153–53.10.8,6=86.153-53.86=86.100=8600
C) 85 .12,7 + 5,3 . 12,7= 12,7(85+5,3)=12,7.90,3=1146,81
D)52.143 – 52 . 39 – 8.26=52(143-39)-8,26=52.104-8,26=5399,74
Bài 1:
a) 5x-20y=5(x-4y)
b) \(5x\left(x-1\right)-3x\left(x-1\right)=2x\left(x-1\right)\)
c) \(x\left(x+y\right)-5x-5y=\left(x+y\right)\left(x-5\right)\)
Bài 2:
a) \(x^2+xy+x\)
\(=x\left(x+y+1\right)\)
\(=77\cdot\left(77+22+1\right)\)
=7700
b) \(x\left(x-y\right)+y\left(y-x\right)\)
\(=x\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)^2\)
\(=50^2=2500\)
Bài 10. Tìm x, biết
a) (x+2)2-x(x+3)+5x=-20 c) (x2-1)3-(x4+x2+1)(x2-1)=0
b) 5x3-10x2+5x=0 d) (x+1)3-(x-1)3-6(x-1)2=-19
Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333
tìm x biết
a) 3x(x-1)+5(x-1)=0
b) x2-3x+2=0
a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Tìm x biết
a) x2 - 25=0
b) x (x + 7) + x + 7=0
a) x= + - 5
b) x\(\in\)\(\left\{-1;-7\right\}\)
a/ \(x^2-25=0\)
\(\Rightarrow\left(x+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\Rightarrow x=-5\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
b/ \(x\left(x+7\right)+x+7=0\)
\(x\left(x+7\right)+\left(x+7\right)=0\)
\(\left(x+7\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
a) x^2 - 25 = 0 <=> (x + 5)(x - 5) = 0
<=> x = -5 hoặc x = 5
b) x(x + 7) + x + 7 = 0 <=> (x + 7)(x + 1) = 0
<=> x = -7 hoặc x = -1
1)Phân tích đa thức thành nhân tử:
a) 11x2-6xy-5y2
b)4x3-16x2+19x-6
2) Tìm x,y biết
a)13x2+y2-16x-6xy+9=0
b)5x2+2y2-4x+6xy+8=0
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
1)Phân tích đa thức thành nhân tử:
a) 11x2-6xy-5y2
b)4x3-16x2+19x-6
2) Tìm x,y biết
a)13x2+y2-16x-6xy+9=0
b)5x2+2y2-4x+6xy+8=0
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-3\right)\left(2x-1\right)\)
\(a,=11x^2-11xy+5xy-5y^2=\left(11x+5y\right)\left(x-y\right)\\ b,=4x^3-8x^2-8x^2+16x+3x-6\\ =\left(x-2\right)\left(4x^2-8x+3\right)\\ =\left(x-2\right)\left(4x^2-2x-6x+3\right)\\ =\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
1)Phân tích đa thức thành nhân tử:
a) 11x2-6xy-5y2
b)4x3-16x2+19x-6
2) Tìm x,y biết
a)13x2+y2-16x-6xy+9=0
b)5x2+2y2-4x+6xy+8=0
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
Bài 2: Tìm đa thức P biết
a)x2+5x+6/x2+4x+4=P/x+2
b)a+1/a-1=(a+1)2/P
c)P/2a-6=a2+3a+9/2
d)a3+b3=(a-b).P
e)x2+y2=(x+y).P
a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)
\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
hay P=x+3
b) Ta có: \(\dfrac{\left(a+1\right)^2}{P}=\dfrac{a+1}{a-1}\)
\(\Leftrightarrow P=\left(a+1\right)\left(a-1\right)\)
\(\Leftrightarrow P=a^2-1\)
Bài 3 :
a) Tìm x biết: (x+2)2 +(x+8)(x+2)
b) Tính giá trị biểu thức : B= (x+y)(x2 – xy + y2) –y3, tại x =10, y = 2021
Bài 3 :
a) Tìm x biết: (x+2)2 +(x+8)(x+2)
b) Tính giá trị biểu thức : B= (x+y)(x2 – xy + y2) –y3, tại x =10, y = 2021
Tìm x,y:
a, x2 + (y -\(\dfrac{1}{10}\))4 = 0
b, (\(\dfrac{1}{2}\) . -5)20 + (y2 - \(\dfrac{1}{4}\))10 ≤0
a: Ta có: \(x^2\ge0\forall x\)
\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)
Do đó: \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(0;\dfrac{1}{10}\right)\)