bài 1 :giải pt
a, 2/x2+2x+1 - 5/x2-2x+1 = 3/1-x3
bài 2 giải pt
a, x2+2x+2/x-1 + x2+8x+20= x2 +4x +6/ x+2 + x2+6x +12/x+3
giúp mình với tối đi hoc rồi
Giải pt : a) 2/-x2+6x-8 - x-1/x-2 = x+3/x-4
b) 2/x3-x2-x+1 = 3/1-x2 - 1/x+1
c) x+2/x-2 - 2/x2-2x = 1/x
d) 5/-x2+5x-6 + x+3/2-x = 0
e) x/2x+2 - 2x/x2-2x-3 = x/6-2x
f) 1/x-1 - 3x2/x3-1 = 2x/x2+x-1
bài 1 :giải pt
a, 2/x2+2x+1 - 5/x2-2x+1 = 3/1-x3
bài 2 giải pt
a, x2+2x+2/x-1 + x2+8x+20= x2 +4x +6/ x+2 + x2+6x +12/x+3
giúp mình với tối đi hoc rồi
Giải pt : a) 2/-x2+6x-8 - x-1/x-2 = x+3/x-4
b) 2/x3-x2-x+1 = 3/1-x2 - 1/x+1
c) x+2/x-2 - 2/x2-2x = 1/x
a,\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\left(đkxđ:x\ne2;4\right)\)
\(< =>\frac{-2}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\frac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)
\(< =>-2-\left(x^2-5x+4\right)=x^2+x-5\)
\(< =>-x^2+5x-6-x^2-x+5=0\)
\(< =>-2x^2+4x-1=0\)
\(< =>2x^2-4x+1=0\)
đến đây thì pt bậc 2 dể rồi
\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\left(đkxđ:x\ne\pm1\right)\)
\(< =>\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}=\frac{3}{1-x^2}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x^2-1\right)\left(x-1\right)}=-\frac{3}{x^2-1}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x+1\right)\left(x-1\right)^2}=\frac{-3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\)
\(< =>2+3x-3+x^2-2x+1=0\)
\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=-1\left(loai\right)\\x=0\left(tm\right)\end{cases}}\)
\(\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\left(đkxđ:x\ne0;x\ne2\right)\)
\(< =>\frac{\left(x+2\right)x}{\left(x-2\right)x}-\frac{2}{x\left(x-2\right)}=\frac{x-2}{x\left(x-2\right)}\)
\(< =>\left(x+2\right)x-2=x-2< =>x^2+2x-x-2+2=0\)
\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=0\left(loai\right)\\x=-1\left(tm\right)\end{cases}}\)
nhớ kết luận tập nghiệm
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
Bài 1: Tính:
a) x2(x-2x3); b) (x2+1)(5-x); c) (x-2)(x2+3x-4); d) (x-2)(x-x2+4); e) (x2-1)(x2+2x); f) (2x-1)(3x+2)(3-x)
Bài 2: Tính:
a) (x-2y)2; b) (2x2+3)3; c) (x-2)(x2+2x+4); d) (2x-1)3
Bài 3: Rút gọn biểu thức:
a) (6x+1)2+(6x-1)2-2(1+6x)(6x-1); b) 3(22+1)(24+1)(28+1)(216+1); c) x(2x2-3)-x2(5x+1)+x2; d) 3x(x-2)-5x(1-x)-8(x2-3)
Bài 4: Tính nhanh:
a) 1012; b) 97.103; c) 772+232+77.46; d) 1052-52; e) A= (x-y)(x2+xy+y2)+2y3 tại x= \(\dfrac{2}{3}\) và y= \(\dfrac{1}{3}\)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
Bài 1: Thu gọn các biểu thức sau
a)(2x2 + 5x - 2)(2x2 - 4x +3)
b)(2x -3)(3x - 2) - 3x(2x - 5)
c)(x -1)(x2 + x + 1) - (x + 1)(x2 - x +1)
d)(x2 + x - 1)(x2 - x + 1)
e)(2 + 3y)2 - (2x -3y)2 -12xy
d)(x2 - 4x)(5 + 2x - x2)
cảm ơn!giúp mình với chiều nay ktra 15ph T_T
Tìm x biết:
a) (x+5).(2x+1)=0
b) x.(x+2)-3.(x+2)=0
c) 2x.(x-5)-x.(3+2x)=26
d) x2-10x-8x+16=0
e) x2-10x=25
f) 5x.(x-1)=x-1
g) 2.(x+5)-x2-5x=0
h) x2+5x-6=0
i) (2x-3)2-4.(x+1).(x-1)=49
j) x3+x2+x+1=0
k) x3-x2=4x2-8x+4
Mn ơi giúp em vs ạ,em cảm ơn trc ạ
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Bài 1 Rút gọn biểu thức
a, [(3x - 2)(x + 1) - (2x + 5)(x2 - 1)] : (x + 1)
b, (2x + 1)2 - 2(2x + 1)(3 - x) + (3 - x)2
c, (x - 1)2 - (x + 1) (x2 - x + 1) - (3x + 1)(1 - 3x)
d, (x2 + 1)(x - 3) - (x - 3)(x2 + 3x + 9)
e, (3x +2)2 + (3x - 2)2 - 2(3x + 2)(3x - 2) + x
Bài 2 Phân tích các đa thức sau thành nhân tử
1, 3(x + 4) - x2 - 4x
2, x2 - xy + x - y
3, 4x2 -25 + (2x + 7)(5 - 2x)
4, x2 + 4x - y2 + 4
5, x3 - x2 - x + 1
6, x3 + x2y - 4x - 4y
7, x3 - 3x2 + 1 - 3x
8, 2x2 + 3x - 5
9, x2 - 7xy + 10y2
10, x3 - 2x2 + x - xy2
Bài 1. Thực hiện phép tính:
a) 2xy(x2+ xy - 3y2)
b) (x + 2)(3x2 - 4x)
c) (x3 + 3x2 - 8x - 20) : (x + 2)
d) (x + y)2 + (x – y)2 – 2(x + y)(x - y) e) (a + b)3 - (a – b)3 – 2b3
f) 2x2(x – 2)+ 3x(x2 – x – 2) –5(3 – x2)
g) (x – 1)(x – 3) – (4 – x)(2x + 1) – 3x2 + 2x – 5
c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)
\(=x^2+x-10\)
Bài 1: Giải các phương trình dưới đây
1) x2 - 9 = (x - 3)(5x +2)
2) x3 - 1 = (x - 1)(x2 - 2x +16)
3) 4x2 (x - 1) - x + 1 = 0
4) x3 + 4x2 - 9x - 36 = 0
5) (3x + 5)2 = (x - 1)2
6) 9 (2x + 1)2 = 4 (x - 5)2
7) x2 + 2x = 15
8) x4 + 5x3 + 4x2 = 0
9) (x2 - 4) - (x - 2)(3 - 2x) = 0
10) (3x + 2)(x2 - 1) = (9x2 - 4) (x + 1)
11) (3x - 1)(x2 + 2) = (3x - 1)(7x - 10)
12) (2x2 + 1) (4x - 3) = (x - 12)(2x2 + 1)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
7.
\(\Leftrightarrow x^2+2x-15=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)
\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)
9.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)
\(\Leftrightarrow x+2=3-2x\)
\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)