tìm x
a) (x-2)3-(x-4)(x2+4x+16)+6(x+1)2
b) (x-2)(x2-2x+4)-x(x2+2)=15
giúp mình với nha
đề bài quy đồng
a, x+2/7x+42; -13x/x2-36
b, 7/4x+16; 15/x2-16
c, 12/x2-4; 2/x-3
d, 2x/x2-1; 5/2x+2
mọi ng giúp e với ạ
Để olm giúp em em nhé!
a, \(\dfrac{x+2}{7x+42}\) = \(\dfrac{x+2}{7.\left(x+6\right)}\) = \(\dfrac{\left(x+2\right)\left(x-6\right)}{7\left(x-6\right)\left(x+6\right)}\) (đk \(x\ne\) \(\mp\) 6)
\(\dfrac{-13x}{x^2-36}\) = \(\dfrac{-13x}{\left(x-6\right)\left(x+6\right)}\) = \(\dfrac{-7.13.x}{7.\left(x-6\right).\left(x+6\right)}\) = \(\dfrac{-91x}{7.\left(x-6\right)\left(x+6\right)}\)
b, \(\dfrac{7}{4x+16}\) = \(\dfrac{7\left(x-4\right)}{4.\left(x+4\right).\left(x-4\right)}\) (đk \(x\ne\) \(\pm\) 4)
\(\dfrac{15}{x^2-16}\) = \(\dfrac{15.4}{\left(x-4\right)\left(x+4\right).4}\) = \(\dfrac{60}{4.\left(x-4\right).\left(x+4\right)}\)
c, \(\dfrac{12}{x^2-4}\) = \(\dfrac{12}{\left(x-2\right).\left(x+2\right)}\) Đk \(x\) \(\ne\) \(\pm\) 2
\(\dfrac{2}{x-3}\) đk \(x\) \(\ne\) 3
\(\dfrac{12}{x^2-4}\) = \(\dfrac{12.\left(x-3\right)}{\left(x^2-4\right).\left(x-3\right)}\) = \(\dfrac{12x-36}{\left(x^2-4\right).\left(x-3\right)}\)
\(\dfrac{2}{x-3}\) = \(\dfrac{2.\left(x^2-4\right)}{\left(x-3\right).\left(x^2-4\right)}\)
Tìm x biết:
a) ( x – 1 ) 3 + ( 2 – x ) ( 4 + 2 x + x 2 ) + 3x(x + 2) = 16;
b) (x + 2)( x 2 – 2x + 4) – x( x 2 – 2) = 15.
a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.
b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .
tìm x, biết:
a) 9x2+36=0
b) 3(x+4)-x2-4x=0
c) x(2x-1)-(x-2)(2x+1)=0
d) (2x-3)2-4x2=00
e)1 phần 3.x2-3x=0
f) x3-x2-x+1=0
ráng giúp mình nha
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
1/Rút gọn các biểu thức: a)(x+1)2-(x-1)2-3(x+1)(x-1)
b)5(x+2)(x-2)-(2x-3)2-x2+17
c)(x-3)3-(x-3)(x2+3x+9)+6(x+1)2
2/ Tìm x a) (x+4)2-(x+1)(x-1)=16
b) (x+2)(x2-2x+4)-x(x2+2)=15
Bài 5: Tìm nghiệm của các đa thức sau: Dạng 1: a) 4x + 9 b) -5x + 6 c) 7 – 2x d) 2x + 5 Dạng 2: a) ( x+ 5 ) ( x – 3) b) ( 2x – 6) ( x – 3) c) ( x – 2) ( 4x + 10 ) Dạng 3: a) x2 -2x b) x2 – 3x c) 3x2 – 4x d) ( 2x- 1)2 Dạng 4: a) x2 – 1 b) x2 – 9 c)– x 2 + 25 d) x2 - 2 e) 4x2 + 5 f) –x 2 – 16 g) - 4x4 – 25 Dạng 5: a) 2x2 – 5x + 3 b) 4x2 + 6x – 1 c) 2x2 + x – 1 d) 3x2 + 2x – 1
1) (\(\dfrac{1}{2}\)x + 3)*(x2- 4x- 6)
2) (6x2 -9x +15)*(\(\dfrac{2}{3}\)x+1)
3) (3x2 -x+5)*(x3+5x-1)
4) (x-1)*(x+1)*(x-2)
5) D=(x-7)*(x+5)-(x-4)*(x+3)
6) E= 4x*(x2-x-1)-(x+3)*(x2-2)
7) F= 5x*(x-3)*(x-1)-4x*(x2-2x)
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)
Bài 1:
a,√3x+4−√2x+1=√x+3
b, √2x−5+√x+2=√2x+1
c, √x+4−√1−x=√1−2x
d, √x+9=5−√2x+4
Bài 2:
a,√x+4√x+4=5x+2
b, √x2−2x+1+√x2+4x+4=4
c, √x+2√x−1+√x−2√x−1=2
d,√x−2+√2x−5+√x+2+3√2x−5=7√2
Bài 3:
a, x2−7x=6√x+5−30
tìm x:
a)3(2x-3)+2(2-x)=-3
b)2x(x2-2)+x2(1-2x)-x2=-12
c)3x(2x+3)-(2x+5)(3x-2)=8
d)4x(x - 1) - 3(x2-5)-x2=(x-3)-(x+4)
e)2(3x-1)(2x+5)-6(2x-1)(x+2)=-6
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Tìm x :
b )(x-1) . ( x2 +x +1) -x.(x-3) . (x+3 )=8
c)( X2 + 2 ) . (x-4 ) - ( X+2 ). ( x2 +4x +4=-16
b) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x-3\right)\left(x+3\right)=8\)
\(\Rightarrow x^3-1-x\left(x^2-9\right)=8\)
\(\Rightarrow x^3-1-x^3+9x=8\)
\(\Rightarrow9x=9\Rightarrow x=1\)
c) \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)\left(x^2+4x+4\right)=-16\)
\(\Rightarrow x^3-4x^2+2x-8-\left(x+2\right)\left(x+2\right)^2=-16\)
\(\Rightarrow x^3-4x^2+2x-8-\left(x+2\right)^3=-16\)
\(\Rightarrow x^3-4x^2+2x-8-\left(x^3+6x^2+12x+8\right)=-16\)
\(\Rightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)
\(\Rightarrow-10x^2-10x-16=-16\)
\(\Rightarrow10x^2+10x=0\)
\(\Rightarrow10x\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
a/ 2b -√b2−4b+4b−2
b/ |x+4| - x+4√x2+8x+16
c/√4−4a+a2−2a với -4 ≤x≤ 2
d/|x+4| - x+4√x2+8x+16
e/√4x^2-4x+1/2x-1với x<1/2
f/|x|+x√x2
với x>0