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a, <=> (x-2)2=25
<=>x-2=5 hoặc x-2=-5
<=>x=7 hoặc x=-3
c,<=>(x2)2-16=0
<=>(x2)2=16
<=>x2=4
<=>x=2 hoặc x=-2
a) Ta có: \(\frac{3x-2}{6}-\frac{4-3x}{18}=\frac{4-x}{9}\)
\(\Leftrightarrow\frac{3\left(3x-2\right)}{18}-\frac{4-3x}{18}-\frac{2\left(4-x\right)}{18}=0\)
\(\Leftrightarrow9x-6-4+3x-\left(8-2x\right)=0\)
\(\Leftrightarrow12x-10-8+2x=0\)
\(\Leftrightarrow10x-18=0\)
\(\Leftrightarrow10x=18\)
hay \(x=\frac{9}{5}\)
Vậy: \(x=\frac{9}{5}\)
b) Ta có: \(\frac{2+3x}{6}-x+2=\frac{x-7}{9}\)
\(\Leftrightarrow\frac{3\left(2+3x\right)}{18}-\frac{18x}{18}+\frac{36}{18}-\frac{2\left(x-7\right)}{18}=0\)
\(\Leftrightarrow6+9x-18x+36-\left(2x-14\right)=0\)
\(\Leftrightarrow42-9x-2x+14=0\)
\(\Leftrightarrow56-11x=0\)
\(\Leftrightarrow11x=56\)
hay \(x=\frac{56}{11}\)
Vậy: \(x=\frac{56}{11}\)
c) ĐKXĐ: x∉{3;-3}
Ta có: \(\frac{6-x}{x^2-9}+\frac{2}{x+3}=\frac{-5}{x-3}\)
\(\Leftrightarrow\frac{6-x}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow6-x+2x-6=-5x-15\)
\(\Leftrightarrow x+5x+15=0\)
\(\Leftrightarrow6x=-15\)
hay \(x=\frac{-5}{2}\)(tm)
Vậy: \(x=\frac{-5}{2}\)
d) Ta có: \(\left(5x+2\right)\left(x^2-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x^2-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-2\\x^2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{5}\\x=\pm\sqrt{7}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-2}{5};\sqrt{7};-\sqrt{7}\right\}\)
e) ĐKXĐ: x∉{4;-4}
Ta có: \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)
\(\Leftrightarrow\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{5x-2}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Leftrightarrow3x+12+5x-2-\left(4x-16\right)=0\)
\(\Leftrightarrow8x+10-4x+16=0\)
\(\Leftrightarrow4x+26=0\)
\(\Leftrightarrow4x=-26\)
hay \(x=\frac{-13}{2}\)(tm)
Vậy: \(x=\frac{-13}{2}\)
a) Đề ( \(x\ne\pm1\))
>\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{4}{\left(x+1\right)\left(x-1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=4\\ \Leftrightarrow2.2x=4\Leftrightarrow x=1\left(kothỏa\right)\)
Vậy \(S=\varnothing\)
b) đề \(\left(x\ne-\frac{1}{2},\frac{1}{2}\right)\)
\(\frac{32x^2}{12\left(1-2x\right)\left(1+2x\right)}=\frac{-8x\left(1+2x\right)}{12\left(1-2x\right)\left(1+2x\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(1+2x\right)}\\ \Leftrightarrow32x^2=-8x-16x^2-3-12x+48x^2\\ \Leftrightarrow20x+3=0\Leftrightarrow x=\frac{20}{3}\left(thỏadk\right)\)
Vậy \(S=\left\{\frac{20}{3}\right\}\)
a, \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
\(=>x+36=0\)
\(=>x=36\)
b. \(\Leftrightarrow\frac{x-1}{99}-1+\frac{x-3}{97}-1+\frac{x-5}{95}-1< \frac{x-2}{98}-1+\frac{x-4}{96}-1+\frac{x-6}{94}-1\)
\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{97}+\frac{x-100}{95}-\frac{x-100}{98}-\frac{x-100}{96}-\frac{x-100}{94}< 0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)< 0\)
Vì \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)< 0\)
Nên \(x-100< 0\Leftrightarrow x< 100\)
P/S: Bài a có sai đề ko bn
Mình nhầm
Vì \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)< 0\)
Nên \(\left(x-100\right)>0\Leftrightarrow x>100\)
\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1=\frac{x-1980}{15}-1+\frac{x-1990}{5}-1\)
\(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)
\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)
Mà \(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\ne0\)
Nên \(x-1995=0\Leftrightarrow x=1995\)
Bài 1:
a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)
\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)
\(\Leftrightarrow-5=0\)(vl)
Vậy: \(x\in\varnothing\)
b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)
hay x=1
Vậy: x=1
c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)
\(\Leftrightarrow2x-72=0\)
\(\Leftrightarrow2\left(x-36\right)=0\)
mà 2>0
nên x-36=0
hay x=36
Vậy: x=36
d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)
\(\Leftrightarrow120x+36=56-64x\)
\(\Leftrightarrow120x+36-56+64x=0\)
\(\Leftrightarrow184x-20=0\)
\(\Leftrightarrow184x=20\)
hay \(x=\frac{5}{46}\)
Vậy: \(x=\frac{5}{46}\)
e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)
\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)
\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)
\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)
\(\Leftrightarrow-23x+29=0\)
\(\Leftrightarrow-23x=-29\)
hay \(x=\frac{29}{23}\)
Vậy: \(x=\frac{29}{23}\)
f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)
\(\Leftrightarrow2x+8-10x-50-25=0\)
\(\Leftrightarrow-8x-67=0\)
\(\Leftrightarrow-8x=67\)
hay \(x=\frac{-67}{8}\)
Vậy: \(x=\frac{-67}{8}\)
g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)
\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)
\(\Leftrightarrow10-5x-8x-8+12x-30=0\)
\(\Leftrightarrow-x-28=0\)
\(\Leftrightarrow-x=28\)
hay x=-28
Vậy: x=-28
h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)
\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(x\in R\)
Bài 2:
a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)
b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)
c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)
\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: Tập nghiệm S={-3}
d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)
\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)
\(\Leftrightarrow12-7x=0\)
\(\Leftrightarrow7x=12\)
hay \(x=\frac{12}{7}\)
Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)
e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x
\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow31x=1\)
hay \(x=\frac{1}{31}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)
Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\)
\(\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\left(2\right)\)
Từ (1) và (2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)
Ta có : \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=\frac{3x}{27}=\frac{2y}{24}=\frac{5z}{50}=\frac{3x-2y+5z}{27-24+50}=\frac{86}{53}\) (đề sai)
b) Đặt : k = \(\frac{x}{5}=\frac{y}{7}\)
=> k2 \(=\frac{x}{5}.\frac{y}{7}=\frac{xy}{35}=\frac{140}{35}=4\)
=> k = -2;2
+ k = 2 thì \(\frac{x}{5}=2\Rightarrow x=10\)
\(\frac{z}{7}=2\Rightarrow z=14\)
+ k = -2 thì \(\frac{x}{5}=2\Rightarrow x=-10\)
\(\frac{z}{7}=2\Rightarrow z=-14\)
Vậy................................
a ) \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{6-5x}{\left(x+2\right)\left(x-2\right)}=\frac{6x-4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}\)
b ) \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{-6x^2+5x-1+6x^2-4x+2-3x}{2x\left(2x-1\right)}=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-1}{2x}\)
c ) \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{\left(x+3\right)^2}+\frac{1}{-\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{-12x+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{x^3-21x}{x^4-18x^2+81}\)
d ) \(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}=\frac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{x^3-1}=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{1}{x^2+x+1}\)
e ) \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x}{x+2y}\)