1, (x-2)(x+2)(x^2+4)-(x^2-3)(x^2+3)
2, (6x+1)^2 +(6x-1)^2 -2(1+6x)(6x-1)
giúp mik với
A= 6x/5x-20 - x/x^2-8x+16
A= 4/x+2 + 3/x-2 + 5x+2/4-x^2 - x^2-2x+4/x^3+8
A= ( 6x+1/x^2-6x) + 6x-1/x^2+6x) . x^2-36/x^2+1
A= ( x/x-1 - x+1/x) : ( x/x+1 - x-1/x)
oke nhé , giúp minh với
\(A=\dfrac{6x}{5x-20}-\dfrac{x}{x^2-8x+16}\)
\(ĐKXĐ:x\ne\pm4\)
\(\Leftrightarrow A=\dfrac{6x}{5\left(x-4\right)}-\dfrac{x}{\left(x-4\right)^2}\)
\(\Leftrightarrow A=\dfrac{6x^2-24x-5x}{5\left(x-4\right)^2}\)
\(\Leftrightarrow\dfrac{6x^2-29x}{5\left(x-4\right)^2}\)
\(\Leftrightarrow\dfrac{x\left(6x-29\right)}{5\left(x-4\right)^2}\)
\(A=\left(\dfrac{x}{x-1}-\dfrac{x+1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\)
\(ĐKXĐ:x\ne0;x\ne\pm1\)
\(\Leftrightarrow A=\left(\dfrac{x^2}{x\left(x-1\right)}-\dfrac{x^2-1}{x\left(x-1\right)}\right):\left(\dfrac{x^2}{x\left(x+1\right)}-\dfrac{x^2-1}{x\left(x+1\right)}\right)\)
\(\Leftrightarrow A=\dfrac{x\left(x+1\right)}{x\left(x-1\right)}\)
\(\Leftrightarrow A=\dfrac{x+1}{x-1}\)
\(A=\left[\dfrac{6x+1}{x^2-6x}+\dfrac{6x-1}{x^2+6x}\right].\dfrac{x^2-36}{x^2+1}\)
\(ĐKXĐ:x\ne0;x\ne\pm6\)
\(\Leftrightarrow A=\left[\dfrac{6x+1}{x\left(x-6\right)}+\dfrac{6x-1}{x\left(x+6\right)}\right].\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(\Leftrightarrow A=\left[\dfrac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\right].\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(\Leftrightarrow A=\left[\dfrac{6x^2+37x+6+6x^2-37x+6}{x\left(x-6\right)\left(x+6\right)}\right].\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(\Leftrightarrow A=\dfrac{12\left(x^2+1\right)}{x\left(x-6\right)\left(x+6\right)}.\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(\Leftrightarrow A=\dfrac{12}{x}\)
A= 6x/5x-20 - x/x^2-8x+16
A= 4/x+2 + 3/x-2 + 5x+2/4-x^2 - x^2-2x+4/x^3+8
A= ( 6x+1/x^2-6x) + 6x-1/x^2+6x) . x^2-36/x^2+1
A= ( x/x-1 - x+1/x) : ( x/x+1 - x-1/x)
oke nhé , giúp minh với
a: \(=\dfrac{6x}{5\left(x-4\right)}-\dfrac{x}{\left(x-4\right)^2}\)
\(=\dfrac{6x^2-24x-5x}{5\left(x-4\right)^2}=\dfrac{6x^2-29x}{5\left(x-4\right)^2}\)
b: \(=\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2-2x+4}{x^3+8}\)
\(=\dfrac{4x-8+3x+6-5x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{1}{x+2}\)
\(=\dfrac{2x-2-x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x}{\left(x+2\right)\left(x-2\right)}\)
c: \(\left(\dfrac{x}{x-1}-\dfrac{x+1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\)
\(=\dfrac{x^2-x^2+1}{x\left(x-1\right)}:\dfrac{x^2-x^2+1}{x\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)}{x\left(x-1\right)}=\dfrac{x+1}{x-1}\)
Bài 3: phân tích thành nhân tử:
1/ 9x^3-xy^2
2/x^2-3xy-6x+18y
3/x^2-3xy-6x+18y 3/6x(x-y)-9y^2+9xy
4/ 6xy-x^2+36-9y^2
5/ x^4-6x^2+5
6/ 9x62-6x-y^2+2y
Bài 4:Tìm x, biết:
1/ (x-1)(x^2+x+1)-x^3-6x=11
2/ 16x^2-(3x-4)^2=0
3/ x^3-x^2+3-3x=0
4/ x-1/x+2=x+2/x+1
5/1/x+2/x+1=0
6/ 9-x^2/x : (x-3)=1
Bài5: 1/ 12x^3y^2/18xy^5
2/10xy-5x^2/2x^2-8y^2
3/ x^2-xy-x+y/x^2+xy-x-y
4/ (x+1)(x^2-2x+1)/(6x^2-6)(x^3-1)
5/ 2x^2-7x+3/1-4x^2
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
1/x-1-x^3-x/x^2+1(x/x^2-2x+1-1/x^2-1)
[2/(x+1)^3.(1/x+1)+1/x^2+2x+1(1/x^2+1)]:x-1/x^3=x/x-1
(x/x^2-36-x-6/x^2+6x):2x-6/x^2+6x+x/6-x
giúp mik với ;-; mik cần gấp
Bài 1. Thu gọn:
a) x2 – 4 – (x + 2)2 | b) (x + 2)(x – 2) – (x – 3)(x + 1) |
c) (x – 2)(x + 2) – (x – 2)(x + 5) | d) (6x + 1)2 + (6x – 1)2 – 2(6x + 1)(6x – 1) |
e) 7a(3a – 5) + (2a -3)(4a + 1) – (6a – 2)2 | g) (5y – 3)(5y + 3) – (5y – 4)2 |
h) (3x + 1)3 – (1 – 2x)3 | i) (2x + 1)2 + 2(4x2 – 1) + (2x – 1)2 |
a: Ta có: \(x^2-4-\left(x+2\right)^2\)
\(=x^2-4-x^2-4x-4\)
=-4x-8
b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-4-x^2+2x+3\)
=2x-1
c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)
\(=\left(x-2\right)\left(x+2-x-5\right)\)
\(=-3x+6\)
d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
=4
e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)
\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)
\(=29a^2-45a-3-36a^2+24a-4\)
\(=-7a^2-21a-7\)
g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)
\(=25y^2-9-25y^2+40y-16\)
=40y-25
h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)
\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)
\(=35x^3+15x^2+15x\)
i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=16x^2\)
1. x^4+x^2-2=0; 2. x^3+3x^2+6x+4=0; 3. x^3-6x^2+8x=0; 4. x^4-8x^3-9x^2=0 Giúp với (;~;)
1/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
2/ \(x^3+3x^2+6x+4=0\)
\(\Leftrightarrow\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow x+1=0\) (do \(x^2+2x+4=\left(x+1\right)^2+3>0,\forall x\))
\(\Leftrightarrow x=-1\).
3/ \(x^3-6x^2+8x=0\)
\(\Leftrightarrow x\left(x^2-6x+8\right)=0\\ \Leftrightarrow x\left[\left(x^2-2x\right)-\left(4x-8\right)\right]=0\\ \Leftrightarrow x\left[x\left(x-2\right)-4\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=4\end{matrix}\right.\)
4/ \(x^4-8x^3-9x^2=0\)
\(\Leftrightarrow x^2\left(x^2-8x-9\right)=0\\ \Leftrightarrow x^2\left(x^2-9x+x-9\right)=0\\ \Leftrightarrow x^2\left(x\left(x-9\right)+\left(x-9\right)\right)=0\\ \Leftrightarrow x^2\left(x+1\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=9\end{matrix}\right.\)
giúp mik câu này với . tìm x : x^4 + 6x^3 + 7x^2 + 6x + 1 = 0
bài 3
a. (6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)
b. x(2x^2-3)-x^2(5x+1)+x^2
c. 3x(x-2)-5x(1-x)-8(x^2-3)
d. 3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
a) Ta có: \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2=2^2=4\)
b) Ta có: \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x-3x^3\)
c) Ta có: \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=24-11x\)
d) Ta có: \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Rút gọn:
A=(x^4 - 3x^2 + 9)(x^2 + 3) + (3 - x^2)^2
M= 5(x+2y)^2 - (3y + 2x)^2 + (4x - y)^2 + 3(x - 2y)(x + 2y)
E= (6x + 1)^2 + (6x - 1)^2 - 2(1 + 6x)(6x - 1)
C= (x^2 + 4)(x + 2)(x -2) - (x^2 - 3)^3
a,\(A=\left(x^4-3x^2+9\right)\left(x^2+3\right)+\left(3-x^2\right)^2\)
\(A=x^6-3x^4+9x^2+3x^4-9x^2+27+9-6x^2+x^4\)
\(A=x^6+x^4-6x^2+36\)
b, \(M=5\left(x+2y\right)^2-\left(3y+2x\right)^2+\left(4x-y\right)^2+3\left(x-2y\right)\left(x+2y\right)\)
\(M=5\left(x^2+4xy+4y^2\right)-\left(9y^2+12xy+4x^2\right)+\left(16x^2-8xy+y^2\right)+3\left(x^2-4y^2\right)\)
\(M=5x^2+20xy+20y^2-9y^2-12xy-4x^2+16x^2-8xy+y^2+3x^2-12y^2\)
\(M=20x^2\)
Các câu còn lại làm tương tự! Chúc bạn học tốt!!!
E=\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(\Leftrightarrow\left(6x+1\right)^2-2\left(1+6x\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(\Leftrightarrow\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\)
\(\Leftrightarrow\left(6x+1-6x+1\right)^2=2^2=4\)