\({2\over x^3 -x^2 -x +1} = {3\over 1 -x^2} - {1\over x -1}\) ;\({x\over x^2 +5x+6}={2\over x^2 +3x+2}\) ;
Những câu hỏi liên quan
Giải các phương trình sau : a, {8 over x-8} + { 11over x-11} {9 over x-9} +{10 over x-10}b, {x over x-3} - {x over x-5} { x over x-4} - { xover x-6}c, { 4over x^2 - 3x + 2 } - { 3 over 2x^2 - 6x +1 } +1 0d, {1over x-1} + {2over x-2} + {3 over x-3} {6 over x-6}e, {2over 2x+1} - {3 over 2x-1} {4over 4x^2 -1}f, { 2xover x +1 } + { 18 over x^2 +2x-3} {2x-5 over x+3}g, {1 over x-1} + { 2x^2 -5 over x^3 -1 } { 4 over x^2 +x+1}
Đọc tiếp
Giải các phương trình sau :
a, \({8 \over x-8} + { 11\over x-11} = {9 \over x-9} +{10 \over x-10}\)
b, \({x \over x-3} - {x \over x-5} = { x \over x-4} - { x\over x-6}\)
c, \({ 4\over x^2 - 3x + 2 } - { 3 \over 2x^2 - 6x +1 } +1 =0\)
d, \({1\over x-1} + {2\over x-2} + {3 \over x-3} = {6 \over x-6}\)
e, \({2\over 2x+1} - {3 \over 2x-1} = {4\over 4x^2 -1}\)
f, \({ 2x\over x +1 } + { 18 \over x^2 +2x-3} = {2x-5 \over x+3}\)
g, \({1 \over x-1} + { 2x^2 -5 \over x^3 -1 } = { 4 \over x^2 +x+1}\)
a, 8/x-8 + 11/x-11 = 9/x-9 + 10/ x-10
b, x/x-3 - x/x-5 = x/x-4 - x/x-6
c, 4/x^2-3x+2 - 3/2x^2-6x+1 +1 = 0
d, 1/x-1 + 2/ x-2 + 3/x-3 = 6/x-6
e, 2/2x+1 - 3/2x-1 = 4/4x^2-1
f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3
g, 1/x-1 + 2x^2 -5/x^3 -1 = 4/ x^2 +x+1
Bài 1: Rút gọn
1) x^2-y^2 over 6x^2y^2
÷ x+y over 12xy
2) 5x over 2x+1
÷ 3x(x-1) over 4x^2-1
3)( 2x-1over 2x+1
-2x-1over 2x+1
) ÷ 4x over 10x-5
4) 2over 9x^2+6x+1
- 3x over 9x^2-1
5) (5over x^2+2x+1 +2x over x^2-1
) ÷ 2x^2+7x-5 over 3x-3
6) (3over x-3 + 2x over x^2-9
+ xover x+3
) ÷ 2xover x+3
7) (3over x^2-9
+1over x^2+3x
-1over x^2-3x
) ÷ x-2over 2x^2+6x
Đọc tiếp
Bài 1: Rút gọn
1) \(x^2-y^2 \over 6x^2y^2 \)÷ \(x+y \over 12xy\)
2) \(5x \over 2x+1 \) ÷ \(3x(x-1) \over 4x^2-1\)
3)( \(2x-1\over 2x+1 \)-\(2x-1\over 2x+1 \)) ÷ \(4x \over 10x-5 \)
4) \(2\over 9x^2+6x+1 \)- \(3x \over 9x^2-1 \)
5) (\(5\over x^2+2x+1 \)+\(2x \over x^2-1 \)) ÷ \(2x^2+7x-5 \over 3x-3\)
6) (\(3\over x-3 \)+ \(2x \over x^2-9 \) + \(x\over x+3 \)) ÷ \(2x\over x+3\)
7) (\(3\over x^2-9 \)+\(1\over x^2+3x \)-\(1\over x^2-3x \)) ÷ \(x-2\over 2x^2+6x\)
1)
ĐK: \(x,y\neq 0\); \(x+y\neq 0\)
\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)
\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)
2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)
\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)
\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)
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3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)
\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)
4) ĐK: \(x\neq \frac{\pm 1}{3}\)
\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)
\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)
\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)
5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)
\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)
\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{3}{(x+1)^2}\)
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6) ĐK: \(x\neq \pm 3\); 0
\(\left(\frac{3}{x-3}+\frac{2x}{x^2-9}+\frac{x}{x+3}\right): \frac{2x}{x+3}\)
\(=\left(\frac{3(x+3)}{(x-3)(x+3)}+\frac{2x}{(x-3)(x+3)}+\frac{x(x-3)}{(x+3)(x-3)}\right). \frac{x+3}{2x}\)
\(=\frac{3(x+3)+2x+x(x-3)}{(x-3)(x+3)}.\frac{x+3}{2x}\)
\(\frac{(x^2+2x+9)(x+3)}{(x-3)(x+3).2x}=\frac{x^2+2x+9}{2x(x-3)}\)
7) ĐK: \(x\neq 2; \pm 3;0\)
\(\left(\frac{3}{x^2-9}+\frac{1}{x^2+3x}-\frac{1}{x^2-3x}\right): \frac{x-2}{2x^2+6x}\)
\(=\left(\frac{3x}{x(x-3)(x+3)}+\frac{x-3}{x(x-3)(x+3)}-\frac{x+3}{(x+3)x(x-3)}\right).\frac{2x(x+3)}{x-2}\)
\(=\frac{3x+x-3-(x+3)}{x(x-3)(x+3)}.\frac{2x(x+3)}{x-2}\)
\(=\frac{3x-6}{x(x-3)(x+3)}.\frac{2x(x+3)}{x-2}=\frac{3(x-2).2x(x+3)}{x(x-3)(x+3)(x-2)}=\frac{6}{x-3}\)
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\(b26 = [{\sqrt{x-1} \over 3\sqrt{x}-1}-{\sqrt{1} \over 3\sqrt{x}+1}{8\sqrt{x} \over 9x-1}]:[1-{3\sqrt{x}-2 \over 3\sqrt{x+1}}]\)
a) rút gọn
a. 2016 : [ 25 - (3x + 2)] 32 . 7b, 52x - 3 - 2 . 52 52 . 3c,{-3 over 4x}-{20 over 11.13}-{20 over 13.15}-{20 over 15.17}-.....-{20 over 53.55}{3 over 11}d,{x over 6}+{x over 10}+{x over 15}+{x over 21}+{x over 28}+{x over 36}+{x over 45}+{x over 55}+{x over 66}+{x over 78}{220 over 39}e, x+(x-1)+(x-2)+(x-3)+......+(x-2016) 2033136
Đọc tiếp
a. 2016 : [ 25 - (3x + 2)] = 32 . 7
b, 52x - 3 - 2 . 52 = 52 . 3
c,\({-3 \over 4x}-{20 \over 11.13}-{20 \over 13.15}-{20 \over 15.17}-.....-{20 \over 53.55}={3 \over 11}\)
d,\({x \over 6}+{x \over 10}+{x \over 15}+{x \over 21}+{x \over 28}+{x \over 36}+{x \over 45}+{x \over 55}+{x \over 66}+{x \over 78}={220 \over 39}\)
e, x+(x-1)+(x-2)+(x-3)+......+(x-2016) = 2033136
tính giá trị các biểu thức:
\(A = {2x^2+5x-3 \over 3x-1}\) lần lượt tại \(x = {1 \over 2}\), \(x = {-1 \over 3}\),\(x = {1 \over 3}\)
\(B = {2x^2-3y^2+1/2 xy \over 3(x+y)}\)tại \(x = {-1 \over 2}\)và y là số nguyên âm lớn nhất
1.Tìm x,biết:
a)\(x+4 \over 2018\)+\(x+3 \over 2019\)+\(x+2 \over 2020\)=-1
b)\(3 \over 5\)+\(2 \over 5\):x=1
c)3/2x-1/-2=4
1. \({3\over 2}x\)+\({1\over 6}\)=\(3{1\over 6}\)
2. x +20% x = 3,6
3. 0,5 x -\({2\over 3}x\)=40%
4. (0,5-\({2\over 3}x\)):\({9\over 10}\)=\({-5\over 3}\)
giúp mình với cac bn ơi
\( {x \over 2012}+{x-1 \over 2013}+{x-2 \over 2014}-{x-3 \over 2015}={x-4 \over 1008}\)
GIẢI PHƯƠNG TRÌNH BẬC NHẤT MỘT ẨN
1. 3(x+2) = 5x+8
2. 2(x-1) = 3(3+x)+3
3. 5-(x-6) = 4(3-2x)
4. \({2x + 3\over 3} + {2x-1\over 6} = 4- {x \over 3}\)
6. \({10x + 3\over 12} =1 + {6 + 8x\over 9}\)
1) 3(x + 2) = 5x + 8
<=> 3x + 6 = 5x + 8
<=> 3x + 6 - 5x - 8 = 0
<=> -2x - 2 = 0
<=> -2x = 0 + 2
<=> -2x = 2
<=> x = -1
2) 2(x - 1) = 3(3 + x) + 3
<=> 2x - 2 = 9 + x + 3
<=> 2x - 2 = 12 + x
<=> 2x - 2 - 12 - x = 0
<=> x - 14 = 0
<=> x = 0 + 14
<=> x = 14
3) 5 - (x - 6) = 4(3 - 2x)
<=> 5 - x + 6 = 12 - 8x
<=> 11 - x = 12 - 8x
<=> 11 - x - 12 + 8x = 0
<=> -1 + 7x = 0
<=> 7x = 0 + 1
<=> 7x = 1
<=> x = 1/7