Tim x, biet:
a.(3-x)+(2x+7)=2x
b. \(\dfrac{x+1}{2}=\dfrac{1}{3}\)
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\(\dfrac{2x-1}{3+x}< 0\Leftrightarrow2+\dfrac{-7}{x+3}< 0\)
Để biểu thức < 0 suy ra \(-\dfrac{7}{x+3}< -2\)\(\Leftrightarrow\dfrac{7}{x+3}>2\Rightarrow x< \dfrac{1}{2}\)
Vậy \(x< \dfrac{1}{2}\)
\(\dfrac{2x-1}{x+3}< 0\)
TH1: 2x-1>0 và x+3<0
=>x>1/2 và x<-3
=>Loại
TH2: 2x-1<0 và x+3>0
=>-3<x<1/2
c: \(\Leftrightarrow2x+2x-6=12-2x\)
=>4x-6=12-2x
=>6x=18
hay x=3
b: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)+x=2x-1\)
\(\Leftrightarrow x^2-1+x=2x-1\)
=>x2-x=0
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)
\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)
\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)
\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)
\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)
\(\Leftrightarrow12x-9=29x-145\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x+136=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\left(tm\right)\)
Vậy \(S=\left\{8\right\}\)
\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)
\(\Rightarrow2x-1=2\left(5-3x\right)\)
\(\Leftrightarrow2x-1=10-6x\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x-11=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{11}{8}\right\}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)
\(\Rightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-5-3x+2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)
\(\Rightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\)
\(2,\dfrac{2x-1}{5-3x}=2\)
\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)
\(\Leftrightarrow4x-5-2x+2+2x=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
lớp 9 gì như lớp 6 thế
a) đề sai
c) <=>x/3 +x/3 -1 =2-x/3
<=>3.x/3 =3 => x=3
b) x<> 0; -2 <=>
x^2 -1 +x =2x-1
<=>x^2 -x =0 => x =0 (l) x =1 nhận
d ; <=> (x+1)/65+1 +(x+3)/63 +1 =(x+5)/61+1 +(x+7)/59+1
<=>(x+66) [1/65+1/63-1/61-1/59] =0
[...] khác 0
x=-66
\(\Leftrightarrow\dfrac{7^x.7^2+7^x.7+7^x}{57}=\dfrac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}\)
\(\Leftrightarrow7^x\left(\dfrac{7^2+7+1}{57}\right)=5^{2x}\left(\dfrac{1+5+5^3}{131}\right)\)
\(\Leftrightarrow7^x\dfrac{57}{57}=5^{2x}\dfrac{131}{131}\Leftrightarrow7^x=5^{2x}\Leftrightarrow7^x=25^x\Leftrightarrow x=0\)
a) 3-x+2x+7=2x
=> 3+x+7=2x
=> 10+x=2x
=> x-2x=-10
=> -x=-10
=> x=10
Vậy x= 10
b) 3(x+1)=2
=> x+1=2/3
=>x=2/3-1
=> x= 2/3 - 3/3
=> x= -1/3
Vậy x = -1/3
Lời giải:
a, \(\left(3-x\right)+\left(2x+7\right)=2x\)
\(\Rightarrow3+x+7=2x\)
\(\Rightarrow x+10=2x\)
\(\Rightarrow x-2x=-10\)
\(\Rightarrow-x=-10\)
\(\Rightarrow x=10\)
b, \(\dfrac{x+1}{2}=\dfrac{1}{3}\)
\(\Rightarrow3.\left(x+1\right)=2.1\)
\(\Rightarrow3\left(x+1\right)=2\)
\(\Rightarrow x+1=\dfrac{2}{3}\)
\(x=\dfrac{-1}{3}\)