tìm x,y biết:
a,x4-x3-7x2+x+6=0
b,2x2+2xy+y2+9=6x-|y+3|
c,(2x2+x)2-4(2x2+x)+3=0
d,(x2+3x+2)(x2+7x+12)=24
giúp mik với,mik cần gấp
Câu 1 (3,0 điểm): Tính
a) 3x2 (2x2 − 5x − 4)
b) (x + 1)2 + ( x − 2 )(x + 3 ) − 4x
c) (6 x5 y2 − 9 x4 y3 +12 x3 y4 ) : 3x3 y2
Câu 2 (4,0 điểm): Phân tích đa thức thành nhân tử
a) 7x2 +14xy b) 3x + 12 − (x2 + 4x)
c ) x2 − 2xy + y2 − z2 d) x2 − 2x −15
Câu 3 (0,5 điểm): Tìm x
a) 3x2 + 6x = 0 b) x (x − 1) + 2x − 2 = 0
Câu 4 (2,0 điểm): Cho hình bình hành ABCD (AB > BC). Tia phân giác của góc D cắt AB ở E, tia phân giác của góc B cắt CD ở F.
a) Chứng minh DE song song BF
b) Tứ giác DEBF là hình gì?
Câu 5 (0,5 điểm ):
Chứng minh rằng A= n3 + (n+1)3 + (n+2)3 chia hết cho 9 với mọi n ∈ N*
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
Bài 2
a) \(7x^2+14xy=7x\left(x+2y\right)\)
b) \(3x+12-\left(x^2+4x\right)=-x^2-x+12=\left(-x+3\right)\left(x+4\right)\)
c) \(x^2-2xy+y^2=\left(x-y\right)^2\)
d) \(x^2-2x-15=x^2+3x-5x-15=\left(x+3\right)\left(x-5\right)\)
a) \(x^3-x^2+3x-3>0\)
\(\Leftrightarrow x^2\left(x-1\right)+3\left(x-1\right)>0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x-1\right)>0\)
Mà: \(x^2+3>0\forall x\)
\(\Leftrightarrow x-1>0\)
\(\Leftrightarrow x>1\)
b) \(x^3+x^2+9x+9< 0\)
\(\Leftrightarrow x^2\left(x+1\right)+9\left(x+1\right)< 0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+1\right)< 0\)
Mà: \(x^2+9>0\forall x\)
\(\Leftrightarrow x+1< 0\)
\(\Leftrightarrow x< -1\)
d) \(4x^3-14x^2+6x-21< 0\)
\(\Leftrightarrow2x^2\left(2x-7\right)+3\left(2x-7\right)< 0\)
\(\Leftrightarrow\left(2x^2+3\right)\left(2x-7\right)< 0\)
Mà: \(2x^2+3>0\forall x\)
\(\Leftrightarrow2x-7< 0\)
\(\Leftrightarrow2x< 7\)
\(\Leftrightarrow x< \dfrac{7}{2}\)
d) \(x^2\left(2x^2+3\right)+2x^2>-3\)
\(\Leftrightarrow2x^4+3x^2+2x^2+3>0\)
\(\Leftrightarrow2x^4+5x^2+3>0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x^2+3\right)>0\)
Mà:
\(x^2+1>0\forall x\)
\(2x^2+3>0\forall x\)
\(\Rightarrow x\in R\)
a: =>x^2(x-1)+3(x-1)>0
=>(x-1)(x^2+3)>0
=>x-1>0
=>x>1
b: =>x^2(x+1)+9(x+1)<0
=>(x+1)(x^2+9)<0
=>x+1<0
=>x<-1
c: 4x^3-14x^2+6x-21<0
=>2x^2(2x-7)+3(2x-7)<0
=>2x-7<0
=>x<7/2
d: =>x^2(2x^2+3)+2x^2+3>0
=>(2x^2+3)(x^2+1)>0(luôn đúng)
a) 3x(x+1)-x(3x+2)
b) 2x(x2-5x+6)+(x-1)(x+3)
c) (x2-xy+y2)-(x2+2xy+y2)
d) (2/5xy+x-y)-(3x+4y)-2/5xy
e) 2xy(x2-4xy+4y2)
f) (x+y)(xy+5)
g) (x3-2x2-x+2):(x-1)
h) (2x2+3x-2):(2x-1)
1. Tìm x,y:
a) (x+2)2 + (x-3)2 = 2x ( x+ 7)
b) x3- 3x2 + 3x - 126 = 0
c) x2 + y2 - 2x + 4y + 5 = 0
d) 2x2 - 2xy + y2 + 4x + 4 = 0
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)
6). – x2 y(xy2 – 1/2 xy + 3/4 x2 y2 )
7). (3xy – x2 + y). 2/3 x2 y
8). (4x3 – 5xy + 2x)( – 1/2 xy)
9). 2x2 (x2 + 3x + 1/2 )
10). – 3/2 x4 y2 (6x4 − 10/9 x2 y3 – y5 )
11). 2 3 x3 (x + x2 – 3/4 x5 )
12). 2xy2 (xy + 3x2 y – 2/3 xy3 )
13). 3x(2x3 – 1/3 x2 – 4x)
14). 3/5 x3 y5 (7x4 + 5x2 y − 10/21 x4 y3 –y4 )
6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
Tìm x biết:
a) 27x3-54x2+36x-8=0
b) x3+15x2+75x+125=0
c) x3-18x2+108x-216=0
d) (x-1)3-x.(x2+3x)=2
e) (x+1)3+(x-2)3-2x2.(x-1,5)=3
Giải chi tiết giúp mình nha.Cảm ơn nhiều
Tìm x biết:
a) 2x2 - 4 = 0
b) (x + 1)2 = 4
c) (2x - 1)2 - 9 = 0
d) x2 - x = 0
a: =>2x^2=4
=>x^2=2
=>\(x=\pm\sqrt{2}\)
b: =>(x+1)^2-4=0
=>(x+1+2)(x+1-2)=0
=>(x+3)(x-1)=0
=>x=1 hoặc x=-3
c: =>(2x-1)^2-3^2=0
=>(2x-1-3)(2x-1+3)=0
=>(2x-4)(2x+2)=0
=>x=2 hoặc x=-1
d: x^2-x=0
=>x(x-1)=0
=>x=0 hoặc x=1
Bài 5: Biết :
a. 3x + 2( 5 - x ) = 0
b. 2x( x + 3 ) + 2( x + 3 ) = 0
Giá trị của x cần tìm là ?
Bài 6: Rút gọn biểu thức:
A = 2x2(-3x3 + 2x2 + x - 1) + 2x(x2 – 3x + 1) giúp mik nhanh dc ko
Thực hiện phép tính:
a,(2x- 4)(x+9)
b,(x2 + 4x +3)(x-2)
c,(x-8)(x+8)
d, x2(7x-5)-7(x3- 4x+6)
e,(x2+2)(x2+x+1)
f,(x2+2)(x4-2x2+4)
g,(x-g)(x+9)
h,(x-2)(2x3-x2+1)+(x2+1)+(x2-2x2)(1-2)x
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Giúp vs ạ
Bài 1 giải các bất phương trình sau
a.x2 - x - 6 = 0
b.2x2 - 7x + 5 < 0
c.3x2 - 9x + 6 ≥ 0
d.2x2 - 5x + 3 < 0
Bài 2 Giải phương trình sau
A.√x2 + x + 5 = √2x2 - 4x + 1
B.√11x2 -14x - 12 = √3x2 + 4x - 7
Bài 2:
a: =>2x^2-4x+1=x^2+x+5
=>x^2-5x-4=0
=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)
b: =>11x^2-14x-12=3x^2+4x-7
=>8x^2-18x-5=0
=>x=5/2 hoặc x=-1/4