a) \(\dfrac{x^2-x}{x-2} \dfrac{4-3x}{x-2}\) b) \(\dfrac{a 2b}{3a-b} \dfrac{2a-5b}{b-3a}\) c) \(\dfrac{2}{x^2-9} \dfrac{1}{x 3}\) d) \(\dfrac{4x}{x^2-4} \dfrac{x}{x 2} \dfrac{2}{x-2}\) e) \(\dfrac{...

3

NN

Nguyễn Ngọc Linh

19 tháng 11 2018

a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)

\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)

NN

Nguyễn Ngọc Linh

19 tháng 11 2018

c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)

Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)

\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)

Giải các pt với tham số là a,b,c a , \(\dfrac{x-a}{3}=\dfrac{x+3}{a}-2\) e, \(3x+\dfrac{x}{a}-\dfrac{3a}{a+1}=\dfrac{4ax}{\left(a+1\right)^2}+\dfrac{\left(2a+1\right)x}{a\left(a+1\right)^2}-\dfrac{3a^2}{\left(a+1\right)^3}\) b, \(\dfrac{x-a}{a+1}+\dfrac{x-1}{a-1}=\dfrac{2a}{1-a^2}\) c, \(\dfrac{x+a-1}{a+2}+\dfrac{x-a}{a-2}+\dfrac{x-a}{4-a^2}\) d,...

XH

Xuân Huy

13 tháng 1 2018

2

DT

Dương Thị Hồng Nhung

13 tháng 1 2018

minh giai phan d, nha bn :

x-a/b+c + x-b/c+a + x-c/a+b=3

=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0

=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0

=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0

Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0

=>x=a+b+c

XH

Xuân Huy

13 tháng 1 2018

g, x - a / b + c + x - b/ c+a + x - c/ a+b = 3x / a+b+c

QN

Quỳnh Như

8 tháng 8 2017

Thực hiện các phép tính sau:
a) \(\dfrac{3x^2-3y^2}{5xy}.\dfrac{15x^2y}{2y-2x}\)

b) \(\dfrac{4a^2-3a+5}{a^3-1}-\dfrac{1-2a}{a^2+a+1}-\dfrac{6}{a-1}\)

c) \(\dfrac{2a^3-2b^3}{3a+3b}.\dfrac{6a+6b}{a^2-2ab+b^2}\)

d) \(x^2+1-\dfrac{x^4+1}{x^2+1}\)

e) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)

3

NL

Nguyễn Lê Thùy Linh

30 tháng 7 2018

e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)

= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)

= \(\dfrac{2x-6}{2x\left(x+3\right)}\)

= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)

NL

Nguyễn Lê Thùy Linh

30 tháng 7 2018

c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)

= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)

= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)

H

__HeNry__

17 tháng 12 2018

rút gọn biểu thức

1) \(\dfrac{a+b}{3a-b}+\dfrac{b}{a+b}-\dfrac{a^2-b^2}{3a-b}\)

2) \(\left(\dfrac{7}{a+b}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right)\div\dfrac{a+1}{2}\)

3) \(\left(x^2+\dfrac{4x^2}{x^2-4}\right)\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x^3-4x}\times\dfrac{x^2-4}{x-2}\right)\)

1

NL

Nguyễn Lê Phước Thịnh

9 tháng 12 2022

2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)

\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)

\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)

3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)

Thực hiện phép tính : a) \(\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{1}{1-x}\) b) \(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\) c) \(\dfrac{4-2x+x^2}{2+x}-2-x\) d) \(\dfrac{3x^2-3y^2}{5xy}\cdot\dfrac{15x^2y}{2y-2x}\) e)...

NH

Nguyễn Hoàng Linh

11 tháng 12 2017

1

PT

Phương Trâm

12 tháng 12 2017

a) \(\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{1}{1-x}\)

\(=\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{-1}{x-1}\)

\(=\dfrac{x^3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{-1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x^4-x+x^3+x+x-1-x+1}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x^4+x^3}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x^3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^3}{x-1}\)

b) \(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)

\(=\dfrac{x^3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x^3\left(x+1\right)-x^2\left(x-1\right)-1\left(x+1\right)+1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^4+x^3-x^3+x^2-x-1+x-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^4+x^2-2}{\left(x-1\right)\left(x+1\right)}\)

c) \(\dfrac{4-2x+x^2}{2+x}-2-x\)

\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{2\left(2+x\right)}{2+x}-\dfrac{x\left(2+x\right)}{2+x}\)

\(=\dfrac{4-2x+x^2-4-2x-2x-x^2}{2+x}\)

\(=\dfrac{-6x}{2+x}\)

Còn lại thì dễ rồi, bạn tự làm nha ^^

Tính:a) \(\dfrac{{3{a^2}}}{{10{b^3}}} \cdot \dfrac{{15b}}{{9{a^4}}}\) b) \(\dfrac{{x - 3}}{{{x^2}}} \cdot \dfrac{{4x}}{{{x^2} - 9}}\)c) \(\dfrac{{{a^2} - 6a + 9}}{{{a^2} + 3a}} \cdot \dfrac{{2a + 6}}{{a - 3}}\) d) \(\dfrac{{x + 1}}{x} \cdot \left( {x + \dfrac{{2 - {x^2}}}{{{x^2} - 1}}}...

B

Buddy

23 tháng 7 2023

1

H9

HT.Phong (9A5)

23 tháng 7 2023

a) \(\dfrac{3a^2}{10b^3}\cdot\dfrac{15b}{9a^4}\)

\(=\dfrac{3a^2\cdot15b}{10b^3\cdot9a^4}\)

\(=\dfrac{1\cdot3}{2\cdot b^2\cdot3\cdot a^2}=\dfrac{3}{6a^2b^2}\)

b) \(\dfrac{x-3}{x^2}\cdot\dfrac{4x}{x^2-9}\)

\(=\dfrac{x-3}{x^2}\cdot\dfrac{4x}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{\left(x-3\right)\cdot4x}{x^2\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{4}{x\left(x+3\right)}\)

c) \(\dfrac{a^2-6x+9}{a^2+3a}\cdot\dfrac{2a+6}{a-3}\)

\(=\dfrac{\left(a-3\right)^2}{a\left(a+3\right)}\cdot\dfrac{2\cdot\left(a+3\right)}{a-3}\)

\(=\dfrac{\left(a-3\right)^2\cdot2\cdot\left(a+3\right)}{a\left(a+3\right)\left(a-3\right)}\)

\(=\dfrac{2\left(a-3\right)}{a}\)

d) \(\dfrac{x+1}{x}\cdot\left(x+\dfrac{2-x^2}{x^2-1}\right)\)

\(=\dfrac{\left(x+1\right)\cdot x}{x}+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{x^2-1}\)

\(=x+1+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{\left(x+1\right)\left(x-1\right)}\)

\(=x+\dfrac{2-x^2}{x\left(x-1\right)}\)

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VL

Vui lòng để tên hiển thị

23 tháng 7 2023

=))) để r xem

H

__HeNry__

25 tháng 2 2019

Câu 1: Rút gọn

a) \(\left(1+\dfrac{a}{x}+\dfrac{a^2}{x^2}\right)\left(1-\dfrac{a}{x}\right)\left(\dfrac{x^2}{a^3-x^3}\right)\)

b) \(\dfrac{x+3a}{2-x}+\dfrac{x-3a}{2+x}-\dfrac{2a}{4-x^2}+a\)

Biết \(x=\dfrac{a}{3x+2}\)

1)Tìm x,y,z biết : x2+y2+z2+\(\dfrac{1}{x^2}+\dfrac{1}{y^2}\)=4 2)Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\)và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\) Tính \(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}\) 3)Cho \(x=\dfrac{a}{3a+2}\).Rút gọn biểu...

0

DT

Duy Trần

21 tháng 8 2017

2

N

ngonhuminh

22 tháng 8 2017

1)

\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)

\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)

N

ngonhuminh

22 tháng 8 2017

dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)

\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)

\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)

AN

Ánh Ngọc Dương

20 tháng 12 2022

Rút gọn

a)\(\dfrac{x}{x+1}+\dfrac{1}{x-1}-\dfrac{2x}{1-x^2}\)

b)\(\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}\)

c)\(\dfrac{2x^2-3x-9}{x^2-9}-\dfrac{x}{x+3}-\dfrac{x+3}{3-x}\)

d)\(\dfrac{x+3}{x-2}+\dfrac{x+2}{1-x}-\dfrac{4x-x^2}{x^2-3x+2}\)

giúp mik vs

cảm ơn <3

BT1: a) 2x-1=0 ; b) 3x-2=5+x ; c) 2(x-3)-4=3(1+x)-5x ; d) \(\dfrac{x+1}{2}\)- \(\dfrac{2x}{3}\)=1 ; e) x(x-2)+3(x-2)=0 ; f) \(\dfrac{x+1}{x-1}\)+ \(\dfrac{3}{x}\)= \(\dfrac{x^2+2}{x^2-x}\)BT2: a) Cho a>b, chứng minh rằng 2a+1>2b-3b) Tìm x để giá trị của biểu thức 3x-1 ≤ giá trị biểu thức x+2c) Giải các bất phương trình sau và biểu diễn tập nghiệm trên trục số (mng giúp mình giải phương trình thôi nha)2x+3>0 ; 3x+1<x-4 ; 2(x+1)+3≥ 3(5-x)...

1

NL

Nguyễn Lê Phước Thịnh

20 tháng 12 2022

a: \(=\dfrac{x^2-x+x+1+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)

b: \(=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)

c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x^2+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x}{x-3}\)

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BJ

Bakaa Jeanne

6 tháng 5 2021