Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
ko bt có sai ko nữa mà mình tìm ra câu a hai nghiệm:\(\frac{-11+\sqrt{69}}{26}\)
và \(\frac{-11-\sqrt{69}}{29}\)
d) \(\frac{1}{2x-3}-\frac{3}{x.\left(2x-3\right)}=\frac{5}{x}\)
\(\Leftrightarrow\frac{x}{x.\left(2x-3\right)}-\frac{3}{x.\left(2x-3\right)}=\frac{5.\left(2x-3\right)}{x.\left(2x-3\right)}\)
\(\Leftrightarrow x-3=5.\left(2x-3\right)\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=\left(-15\right)+3\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow9x=12\)
\(\Leftrightarrow x=12:9\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{4}{3}\right\}.\)
Chúc bạn học tốt!
a ) \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{6-5x}{\left(x+2\right)\left(x-2\right)}=\frac{6x-4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}\)
b ) \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{-6x^2+5x-1+6x^2-4x+2-3x}{2x\left(2x-1\right)}=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-1}{2x}\)
c ) \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{\left(x+3\right)^2}+\frac{1}{-\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{-12x+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{x^3-21x}{x^4-18x^2+81}\)
d ) \(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}=\frac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{x^3-1}=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{1}{x^2+x+1}\)
e ) \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x}{x+2y}\)
a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)
\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)
\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)
\(=-2x^2+2x+6\)
\(=-2\left(x^2-x-3\right)\)
b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)
\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
\(=4\left(x^2+5\right)\)
c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)
\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)
\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)
\(=-7x^2-20xy-17y^2+1\)
d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
\(=-3x^2\left(x^2-1\right)\)
\(=-3x^2\left(x-1\right)\left(x+1\right)\)
e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2\)
\(=\left(-2\right)^2=4\)
g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y+z\right)^2\)
\(=\left(x+2z\right)^2\)
h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)
\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)
\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)
\(=5x^2+2x^2+3x-1-3x-3\)
\(=7x^2-4\)
Bài 1:
\(x^2+x-6=x^2+3x-2x+6\)
\(=x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
\(b,x^4+2x^3+x^2=\left(x^2+x\right)^2\)
\(e,x^2+5x-6=x^2+6x-x-6\)
\(=x\left(x+6\right)-\left(x+6\right)=\left(x-1\right)\left(x+6\right)\)
\(f,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)\(g,7x-6x^2-2=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2-3x\right)\left(2x-1\right)\)\(i,2x^2+3x-5=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
\(j,16x-5x^2-3=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)=\left(5x-1\right)\left(x+3\right)\)
Bài 2,
\(a,5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
\(b,2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
1,Thực hiện phép tính :
a, (x + 2)9 : (x + 2)6
=(x+2)9-6
=(x+2)3
b, (x - y) 4 : (x - 2)3
=(x-y)4-3
=x-y
c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)
=(x2+2x+4)5-1
=(x2+2x+4)4
d, 2(x2 + 1)3 : 1/3(x2 + 1)
=(2÷1/3).[(x2+1)3÷(x2+1)]
=6(x2+1)2
e, 5 (x - y)5 : 5/6 (x - y)2
=(5÷5/6).[(x-y)5÷(x-y)2]
=6(x-y))3
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
Bài 1:
a) Ta có: \(\left(x^2-2x+1\right):\left(x-1\right)\)
\(=\left(x-1\right)^2:\left(x-1\right)\)
=x-1
b) Ta có: \(\left(x^3+1\right):\left(x^2-x+1\right)\)
\(=\frac{\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}=x+1\)
c) Ta có: \(\left(x^3-x^2-5x-3\right):\left(x-3\right)\)
\(=\frac{x^3-3x^2+2x^2-6x+x-3}{x-3}\)
\(=\frac{x^2\left(x-3\right)+2x\left(x-3\right)+\left(x-3\right)}{\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(x^2+2x+1\right)}{\left(x-3\right)}\)
\(=\left(x+1\right)^2\)
d) Ta có: \(\left(x^4+x^3-6x^2-5x+5\right):\left(x^2+x-1\right)\)
\(=\frac{x^4+x^3-x^2-5x^2-5x+5}{x^2+x-1}\)
\(=\frac{x^2\left(x^2+x-1\right)-5\left(x^2+x-1\right)}{x^2+x-1}\)
\(=\frac{\left(x^2+x-1\right)\left(x^2-5\right)}{x^2+x-1}\)
\(=x^2-5\)