Tìm x, biết:

a) x – 3 = (...">

K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

21 tháng 5 2021

2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)

Khi |x - 1| = 2

=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)

Khi x = - 1 (không thỏa mãn) => Không tìm được A 

b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)

Đẻ P < 8

=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)

=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)

Vậy x > - 1 thì P < 8 

21 tháng 5 2021

Thay x = 1/2 vào 

12 tháng 4 2019

:>>>>>

làm nhiều rồi 

hehe

hihi

30 tháng 8 2019

3/

a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)

\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)

\(A=x^2-2xy+y^2+x^2+2xy+y^2\)

\(A=2x^2+2y^2\)

b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)

\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)

\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)

\(B=8ab\)

c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)

\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)

\(C=x^2+2xy+y^2-x^2+2xy-y^2\)

\(C=4xy\)

d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)

\(D=4x^2-4x+1-8x^2+24x-18+4\)

\(D=-4x^2+20x-13\)

29 tháng 9 2019

a. \(8x\left(x-2017\right)-2x+4034=0\)

\(8x\left(x-2017\right)-2\left(x-2017\right)=0\)

\(\left(8x-2\right)\left(x-2017\right)=0\)

\(\Rightarrow TH1:8x-2=0\)

\(8x=2\)

\(x=\frac{1}{4}\)

\(TH2:x-2017=0\)

\(x=2017\)

Vậy \(x\in\left\{\frac{1}{4};2017\right\}\)

29 tháng 9 2019

Bài 1 

a) \(8x\left(x-2017\right)-2x+4034=0\)

\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)

1 tháng 9 2020

a) x3 + 3x2 + 3x + 1 = 64

=> (x + 1)3 = 64

=> (x + 1)3 = 43

=> x + 1 = 4 => x = 3

b) x3 + 6x2 + 9x = 4x

=> x3 + 6x2 + 9x - 4x = 0

=> x3 + 6x2 + 5x = 0

=> x3 + 5x2 + x2 + 5x = 0

=> x2(x + 5) + x(x + 5) = 0

=> (x + 5)(x2 + x) = 0

=> (x + 5)x(x + 1) = 0

=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)

c) 4(x - 2)2 = (x + 2)2

=> 4(x2 - 4x + 4) = x2 + 4x + 4

=> 4x2 - 16x + 16 = x2 + 4x + 4

=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0

=> 3x2 - 20x + 12 = 0

=> 3x2 - 18x - 2x + 12 = 0

=> 3x(x - 6) - 2(x - 6) = 0

=> (x - 6)(3x - 2) = 0

=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)

d) x4 - 16x2 = 0

=> x2(x2 - 16) = 0

=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)

e) x4 - 4x3 + x2 - 4x = 0

=> x4 + x2 - 4x3 - 4x = 0

=> x2(x2 + 1) - 4x(x2 + 1) = 0

=> (x2 - 4x)(x2 + 1) = 0

=> x(x - 4)(x2 + 1) = 0

=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)

f) x3 + x = 0 => x(x2  + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)

1 tháng 9 2020

\(a,x^3+3x^2+3x+1=64\)

\(\left(x+1\right)^3=64\)

\(\left(x+1\right)^3=4^3\)

\(x+1=4\)

\(x=3\)

\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)

\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)

\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)

\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)

\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)

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

b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)

thay \(x=-\frac{3}{2}\)  vào P tâ đc:   \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)

c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)

\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)

\(\Leftrightarrow4x+2=x^2-3x\)

\(\Leftrightarrow x^2-7x-2=0\)

\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)

bạn tự giải nốt nhé!!

d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)

\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

bạn tự làm nốt nhé

9 tháng 3 2022

a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)

\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)

b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)

12 tháng 3 2020

\(a.\frac{7x-3}{x-1}=\frac{2}{3}\\\Leftrightarrow \frac{3\left(7x-3\right)}{3\left(x-1\right)}= \frac{2\left(x-1\right)}{3\left(x-1\right)}\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\\Leftrightarrow 3\left(7x-3\right)-2\left(x-1\right)=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\frac{7}{19}\)

\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\\ \Leftrightarrow\frac{4\left(3-7x\right)}{2\left(1+x\right)}=\frac{1\left(1+x\right)}{2\left(1+x\right)}\\\Leftrightarrow 4\left(3-7x\right)=1\left(1+x\right)\\ \Leftrightarrow4\left(3-7x\right)-1\left(1+x\right)=0\\ \Leftrightarrow12-28x-1-x=0\\ \Leftrightarrow11-29x=0\\ \Leftrightarrow-29x=-11\\ \Leftrightarrow x=\frac{-11}{-29}=\frac{11}{29}\)

\(c.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\\ \Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x+2\right)\left(3x-1\right)}\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)-\left(5x-7\right)\left(3x+2\right)=0\\ \Leftrightarrow15x^2-5x-3x+1-15x^2-10x+21x+14=0\\ \Leftrightarrow3x+15=0\\\Leftrightarrow 3x=-15\\\Leftrightarrow x=-5\)

\(d.\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\\\Leftrightarrow \frac{\left(4x+7\right)\left(3x+4\right)}{\left(x-1\right)\left(3x+4\right)}=\frac{\left(12x+5\right)\left(x-1\right)}{\left(3x+4\right)\left(x-1\right)}\\\Leftrightarrow \left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow \left(4x+7\right)\left(3x+4\right)-\left(12x+5\right)\left(x-1\right)=0\\ \Leftrightarrow12x^2+16x+21x+28-12x^2-12x+5x-5=0\\ \Leftrightarrow30x+23=0\\ \Leftrightarrow30x=-23\\ \Leftrightarrow x=\frac{-23}{30}\)

\(e.\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\\ \Leftrightarrow1+3\left(x-2\right)=3-x\\\Leftrightarrow 1+3x-6=3-x\\\Leftrightarrow 1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\)

12 tháng 3 2020

\(f.\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8\left(x-7\right)}{x-7}=\frac{1}{x-7}\\ \Leftrightarrow8-x-8\left(x-7\right)=1\\ \Leftrightarrow8-x-8\left(x-7\right)-1=0\\\Leftrightarrow 8-x-8x+56-1=0\\\Leftrightarrow 63-9x=0\\\Leftrightarrow -9x=-63\\ \Leftrightarrow x=\frac{-63}{-9}=7\)

\(g.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\\Leftrightarrow \frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\\\Leftrightarrow \left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)-20=0\\ \Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\)

\(j.\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\\\Leftrightarrow \frac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{2.2x}{2\left(x+1\right)\left(x-3\right)}\\ \Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\\\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)-4x=0\\\Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right. \)

18 tháng 10 2020

a) ( 3 - x )( x2 + 2x - 7 ) + ( x - 3 )( x2 + x - 5 )

= ( 3 - x )( x2 + 2x - 7 ) - ( 3 - x )( x2 + x - 5 )

= ( 3 - x )( x2 + 2x - 7 - x2 - x + 5 )

= ( 3 - x )( x - 2 )

b) ( x - 5 )2 + 3( 5 - x )

= ( x - 5 )2 - 3( x - 5 )

= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )

c) 2x( x - 1 )2 - ( 1 - x )3

= 2x( 1 - x )2 - ( 1 - x )3

= ( 1 - x )2( 2x - 1 + x ) = ( 1 - x )2( 3x - 1 )

d) x2 + 8x + 16 = ( x + 4 )2

e) x2 - 4xy + 4y2 = ( x - 2y )2

g) 4x2 - 25y2 = ( 2x )2 - ( 5y )2 = ( 2x - 5y )( 2x + 5y )

h) 25( x + 1 )2 - 4( x - 3 )2

= 52( x + 1 )2 - 22( x - 3 )2

= ( 5x + 5 )2 - ( 2x - 6 )2

= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )

= ( 3x + 11 )( 7x - 1 )

i) x3 + 27 = ( x + 3 )( x2 - 3x + 9 )

k) 8x3 - 125 = ( 2x )3 - 53 = ( 2x - 5 )( 4x2 + 10x + 25 )

l) x3 + 6x2 + 12x + 8 = ( x + 2 )3

m) -x3 + 9x2 - 27x + 27 = -( x3 - 9x2 + 27x - 27 ) = -( x - 3 )3