Tìm x,y,z
a) (x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15
b) (x^2-2)^2+4(x-1)^2-4(x^2-2)(x-1)=0
c) x^2+y^2+z^2= 4x-2y+6z-14
d) 8x^3-12x^2+6x-1=0
e) x^2+5y^2-4xy-8y+2x+5=0
f) x(x-5)(x+5)-(x-2)(x^2+2x+4)=3
Bài 3 : Tìm x biết
a) (x-2)^2-x(x-3)=0
b) (x+3)(2x+1)-2(x-1)^2=0
c) (4x-5)^2=9(2-5x)^2
d) X^2-6x-13=0
e) (x+2)(x^2-2x+4)-x(x^2+2)=15
f) X^3-6x^2+12x-19=0
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
B1: quy đồng mẫu số các phân thức:
a. 5/ 6x^2y ; 7/ 12xy^2 ; 11/ 18xy
b. 4x+2/ 15x^3y ; 5y - 3/ 9x^2y ; x+1/5xy^3
c. 3/2x ; 3x-3/2x-1 ; 3x-2/2x- 4x^2
d. x^3 + 2x / x^3+1 ; 2x/ x^2 - x +1 ; 1/ x+1
e. y/ 2x^2 - xy ; 4x/ y^2 - 2xy
f. 1/x+2 ; 3/ x^2 - 4 ; x-14/ ( x^2 + 4x + 4 ) (x-2)
g. 1/x+2 ; 1/ (x+2)(4x+7) ;
h. 1/x+3 ; 1/ (x+3)(x+2) ; 1/ (x+2)(4x+7)
B2: dùng quy tắc đổi dấu để tìm mẫu thức chung :
a.4/ x+2 ; 2/x-2 ; 5x-6/4-x^2
b. 1-3x/2x ; 3x-2/2x-1 ; 3x-2/2x-4x^2
c. 1/ x^2 + 6x + 9 ; 1/ 6x-x^2-9 ; x/ x^2 -9
d. x^2 + 2/ x^3 - 1 ; 2/ x^2 + x +1 ; 1/ 1-x
e. x/ - 2y ; x/ x+2y ; 4xy/ 4y^2 - x^2
Ai làm xong trước mình tick nha!
a)9x2 – 49 = 0
b)(x – 1)(x + 2) – x – 2 = 0
c)(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7
d)x(3x + 2) + (x + 1)2 – (2x – 5)(2x + 5) = 0
e)(x + 3)(x2 – 3x + 9) –x(x – 1)(x + 1) – 27 = 0
f)(4x-3)^2-3x(3-4x)=0
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
Bài 1:Rút gọn biểu thức
a.(x-2)(2x-1)-(2x-3)(x-1)-2
b. x(x+3y+1) -2y (x-1) - (y+x+1)x
Bài 2: Tìm x
a. (14x^3 + 12x^2 -14x) :2x = (x+2) (3x-4)
b. (4x - 5) (6x+1) - (8x+3) (3x-4) =15
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
tìm x biết:
a) x - 3= (3 - x)^2
b) x^3 + 3/2x^2 + 3/4x + 1/8 = 1/64
c) 8x^3 - 50x = 0
d) (x - 2) (x^2 + 2x + 7) + 2(x^2 - 4) - 5(x - 2) = 0
e) x(x + 3) - x^2 - 3x = 0
f) x^3 + 27 + (x + 3) (x - 9) = 0
Giúp mik vs ạ
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
Bài 1: Viết thành tích
1: 81-100n^8.
2:m^2 -25
3:16x^2 -25y^2
4: 100a^2b^4 -49
5:8x^3 -27y^6
6: 64 + 125a^6
7: x^6 -y^6
8: -8y^9 -x^6
9:1/4x^4y^2 -3x^5y +9x^6
10:x^2 -2xy -16 +y^2
11: x^2 -6x -y^2 +9
12:x^2 -9 +4xy + 4y^2
Bài 2: Phân tích đa thức thành nhân tử
a, x^4 +2x^3 +x^2
b, x^3 -x +3x^2y + 3xy^2 +y^3 -y
c, 5x^2 -10 xy +5y^2 -20 z^2
d, x^2 +5x -6
e, 5x^2+5xy - x - y
f, 7x -6x^2 -2
g, x^2 +4x +3
h, 2x^2 +3x -5
i, 16x - 5x^2 -3
j, (x^2 +x)^2 +1 (x^2 +x) -12
k, (x^2 +x +1) . (x^2 +x +1) -12
I, (x+1) . (x+2) .(x+3) .(x+4) -24
Bài3:Tìm x biết
1, (x+3) (x^2 -3x +9) - x(x^2-3) = 8(5 -x)
2, (2x +1)^3 +(2x+3)^3 = 0
3, (5x +1)^2 - (5x -3) (5x +3) = 30
4, ( x+3) (x^2 -3x+9) - x(x-2) (x+2) = 15
Mọi người làm giúp mình vs mai mình nộp r
1)6x^2-12x
2) x^2+2x+1-y^2
3) x+y+z+x^2+xy+xz
4)xy+xz+y^2+yz
5)x^3+x^2+x+1
6)xy+y-2x-2
7)x^3+3x-3x^2-9
8)x^2+2xy+x+2y
9) x^2-y^2-2x-2y
10) 7x^2-7xy-5x=5y
a) 6x2 - 12x
= 6x(x - 2)
b) x2 + 2x + 1 - y2
= (x2 + 2x + 1) - y2
= (x + 1)2 - y2
= (x + 1 - y)(x + 1 + y)
c) x + y + z + x2 + xy + xz
= (x + x2) + (y + xy) + (z + xz)
= x(1 + x) + y(1 + x) + z(1 + x)
= (x + y + z)(x + 1)
d) xy + xz + y2 + yz
= (xy + xz) + (y2 + yz)
= x(y + z) + y(y + z)
= (x + y)(x + z)
e) x3 + x2 + x + 1
= (x3 + x2) + (x + 1)
= x2(x + 1) + (x + 1)
= (x2 + 1)(x + 1)
f) xy + y - 2x - 2
= (xy + y) - (2x + 2)
= y(x + 1) - 2(x + 1)
= (y - 2)(x + 1)
g) x3 + 3x - 3x2 - 9
= (x3 - 3x2) + (3x - 9)
= x2(x - 3) + 3(x - 3)
= (x2 + 3)(x - 3)
h) x2 - y2 - 2x - 2y
= (x2 - y2) - (2x + 2y)
= (x + y)(x - y) - 2(x + y)
= (x + y)(x - y - 2)
i) 7x2 - 7xy - 5x = 5y
mk thấy con này sai sai ý
i) 7x2 - 7xy - 5x + 5y
= (7x2 - 7xy) - (5x - 5y)
= 7x(x - y) - 5(x - y)
= (7x - 5)(x - y)
1. (2x+3y)^2
2. (5x-y)^2
3. (2x+y^2)^3
4. (2x-1) (4x^2+2x+1)
5. (5+3x)^3
6.(x^2+2/5y) (x^2-2/5y)
7.(x+1/4)^2
8. (2/3x^2-1/2y)^3
9. (3x^2-2y)^3
10.(x-3y) (x^2+3xy+9y^2)
11. (x^2-3) (x^4+3x^2+9)
12. (x+2y+z) (x+2y-z)
mn giúp em với