3x-2/3x-1=x+1/x-1
Cho R(x) = 2x 2 + 3x - 1; M(x) = x 2 - x 3 thì R(x) - M(x)=
A.-3x 3 + x 2 + 3x – 1 B. -3x 3 - x 2 + 3x – 1
B. 3x 3 - x 2 + 3x – 1 D. x 3 + x 2 + 3x + 1
R(x) = 2x2 + 3x - 1
- M(x) = -x3 + x2
x3 + x2 + 3x - 1
Vậy R(x) - M(x) = x3 + x2 + 3x - 1
Rút gọn các biểu thức sau:
a,(3x+1)^2-2(3x+1)(3x-5)+(3x-5)^2
b,(3x^2-y)^2
c,(3x+5)^2+(3x-5)^2-(3x+2)(3x-2)
d,2x(2x-1)^2-3x(x+3)(Õ-3)-4x(x+1)^2
e,(x-2)(x^2+2x+4)-(x+1)^2+3(x-1)(x+1)
f,(x^4-5x^2+25)(x^2+5)-(2+x^2)^2+3(1+x^2)^2
a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
b1:
[4+2x]=-3x [3x-1]+2=x
[x+15]+1=3x [2x-5]+x=2
b2:
[2x-5]=x+1 [3x-2]-1=x
[3x-7]=2x+1 [2x-1]+1=x
1.Giải phương trình:
a) 4x-8/2x^2+1 = 0
b)x^2-x-6/x-3 = 0
c)x+5/3x-6 - 1/2 = 2x-3/2x-4
d)12/1-9x^2 = 1-3x/1+3x - 1+3x/1-3x
2.Giải các phương trình:
a)5 + 96/x^2-16 = 2x-1/x+4 - 3x-1/4-x
b)3x+2/3x-2 - 6/2+3x = 9x^2/9x^2-4
c)x+1/x^2+x+1 - x-1/x^2-x+1 = 3/x(x^4+x^2+1)
Bài 1.
\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)
Bài 2.
\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)
ĐK: \(x\ne2\)
\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)
ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
Bài 2.
\(a)5 + \dfrac{{96}}{{{x^2} - 16}} = \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{4 - x}}\)
ĐK: \(x\ne\pm4\)
\( Pt \Leftrightarrow \dfrac{{96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} - \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{x - 4}} = - 5\\ \Leftrightarrow \dfrac{{96 - \left( {2x - 1} \right)\left( {x - 4} \right) - \left( {3x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow \dfrac{{ - 5{x^2} - 2x + 96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow - 5{x^2} - 2x + 96 = - 5\left( {{x^2} - 16} \right)\\ \Leftrightarrow 96 - 2x = 80\\ \Leftrightarrow - 2x = - 16\\ \Leftrightarrow x = 8\left( {tm} \right)\\ b)\dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} = \dfrac{{9{x^2}}}{{9{x^2} - 4}} \)
ĐK: \(x \ne \dfrac{2}{3};x \ne -\dfrac{2}{3}\)
\( Pt \Leftrightarrow \dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} - \dfrac{{9{x^2}}}{{9{x^2} - 4}} = 0\\ \Leftrightarrow \dfrac{{{{\left( {2 + 3x} \right)}^2} - 6\left( {3x - 2} \right) - 9{x^2}}}{{\left( {3x - 2} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{16 - 6x}}{{\left( {3 - 2x} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow 16 - 6x = 0\\ \Leftrightarrow - 6x = - 16\\ \Leftrightarrow x = \dfrac{8}{3}\left( {tm} \right)\\ c)\dfrac{{x + 1}}{{{x^2} + x + 1}} - \dfrac{{x - 1}}{{{x^2} - x + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}} \)
Ta có: \(x(x^4+x^2+1)=x[(x^2+1)^2-x^2]=x(x^2+x+1)(x^2-x+1)\)
Do \(\left\{ \begin{array}{l} {x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x\\ {x^2} - x + 1 = \left( {x - \dfrac{1}{2}} \right) + \dfrac{3}{4} > 0\forall x \end{array} \right.\) nên phương trình xác định với mọi $x \ne 0$
Quy đồng, rồi biến đổi phương trình về dạng \(2x=3 \Leftrightarrow x =\dfrac{3}{2} (tm)\)
a) x\(^2\)-3x+7=1+2x
b) x\(^2\)-3x-10=0
c) x\(^2\)-3x+4=2(x-1)
d) (x+1)(x-2)(x-5)=0
e) 2x\(^2\)+3x+1=0
f) 4x\(^2\)-3x=2x-1
a) Ta có: \(x^2-3x+7=1+2x\)
\(\Leftrightarrow x^2-3x+7-1-2x=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: S={3;2}
b) Ta có: \(x^2-3x-10=0\)
\(\Leftrightarrow x^2-5x+2x-10=0\)
\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy: S={5;-2}
c) Ta có: \(x^2-3x+4=2\left(x-1\right)\)
\(\Leftrightarrow x^2-3x+4=2x-2\)
\(\Leftrightarrow x^2-3x+4-2x+2=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: S={3;2}
d) Ta có: \(\left(x+1\right)\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=5\end{matrix}\right.\)
Vậy: S={-1;2;5}
e) Ta có: \(2x^2+3x+1=0\)
\(\Leftrightarrow2x^2+2x+x+1=0\)
\(\Leftrightarrow2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{-1}{2}\right\}\)
f) Ta có: \(4x^2-3x=2x-1\)
\(\Leftrightarrow4x^2-3x-2x+1=0\)
\(\Leftrightarrow4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{1}{4}\right\}\)
Giải ptrinh :
\(\dfrac{x^2}{\sqrt{3x-2}}-\sqrt{3x-2}=1-x\)
\(\sqrt{x+1}+2\left(x+1\right)=x-1+\sqrt{1-x}+3\sqrt{1-x^2}\)
\(3x^2+3x+2=\left(x+6\right)\sqrt{3x^2-2x-3}\)
tính
\(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)
\(\dfrac{3x+1}{3x^2-3x+1}+\dfrac{x^2-6x}{x^2-3x+1}\)
\(\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)
a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)
\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)
c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)
\(=\dfrac{-x+7x-4}{3x-2}\)
\(=\dfrac{6x-4}{3x-2}=2\)
\(1, (-x+5)(x-2)+(x-7)(x+7)=(3x+1)^2-(3x-2)(3x+2)\)
\(2, (5x-1)(x+1)-2(x-3)^2=(x+2)(3x-1)-(x+4)^2+(x^2-x)\)
3, \((x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2 +(x-2)^2-x^2\)
1: \(A=\left(-x+5\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)\)
\(=-x^2+2x+5x-10+x^2-49=7x-59\)
\(B=\left(3x+1\right)^2-\left(3x-2\right)\left(3x+2\right)\)
\(=9x^2+6x+1-9x^2+4=6x+5\)
=>7x-59=6x+5
=>x=64
2: \(A=\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2\)
\(=5x^2+5x-x-1-2x^2+12x-9\)
\(=3x^2+16x-10\)
\(B=\left(x+2\right)\left(3x-1\right)-\left(x+4\right)^2+x^2-x\)
\(=3x^2-x+6x-2-x^2-8x-16+x^2-x\)
\(=3x^2-4x-18\)
=>16x-10=-4x-18
=>20x=-8
hay x=-2/5
Rút gọn biểu thức:
a) (x 2 – 2x + 2)(x 2 – 2)(x 2 + 2x + 2)(x 2 + 2)
b) (x + 1)2 – (x – 1)2 + 3x 2 – 3x(x + 1)(x – 1)
c) (2x + 1)2 + 2(4x 2 – 1) + (2x – 1)2
d) (m + n)2 – (m – n)2 + (m – n)(m + n)
e) (3x + 1)2 – 2(3x + 1)(3x + 5) + (3x + 5)2
a: Ta có: \(\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
\(=\left(x^4-4\right)\left[\left(x^2+2\right)^2-4x^2\right]\)
\(=\left(x^4-4\right)\left(x^4+4x^2+4-4x^2\right)\)
\(=\left(x^4-4\right)\cdot\left(x^4+4\right)\)
\(=x^8-16\)
b: Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2+3x^2-3x\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-x^2+2x-1+3x^2-3x\left(x^2-1\right)\)
\(=3x^2+4x-3x^3+3x\)
\(=-3x^3+3x^2+7x\)