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\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
\(\Leftrightarrow A=\left(\dfrac{x}{x+2}+\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{\left(x-2\right)^2}{-\left(x-2\right)\left(x+2\right)}\right):\dfrac{4}{x+2}\)
\(\Leftrightarrow A=\left(\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)^2}.\dfrac{-\left(x-2\right)}{\left(x+2\right)}\right):\dfrac{4}{x+2}\)
\(\Leftrightarrow A=\left(\dfrac{x}{x+2}-1\right):\dfrac{4}{x+2}\)
\(\Leftrightarrow A=\dfrac{2}{x+2}:\dfrac{4}{x+2}\)
\(\Leftrightarrow A=\dfrac{1}{2}\)
\(A=\left(\dfrac{x}{x+2}+\dfrac{x^3-8}{x^3+8}.\dfrac{x^2-2x+4}{4-x^2}\right):\dfrac{4}{x+2}=\left(\dfrac{x}{x+2}+\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{-\left(x-2\right)\left(2+x\right)}\right).\dfrac{x+2}{4}=\left(\dfrac{x\left(x+2\right)}{\left(x+2\right)^2}-\dfrac{\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right).\dfrac{x+2}{4}=\left(\dfrac{x^2+2x-x^2-2x-4}{\left(x+2\right)^2}\right).\dfrac{x+2}{4}=\dfrac{-4}{\left(x+2\right)^2}.\dfrac{x+2}{4}=-\dfrac{1}{x+2}\)
a) 3x+2(x-5)=-x+2
<=> 3x+2x+x=2+10
<=>6x=12
<=>x=2
b) 3x2-2x=0
<=>x(3x-2)=0
<=>\(\left[{}\begin{matrix}x=0\\3x-2=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) \(\dfrac{2x}{3}\)+\(\dfrac{x-4}{6}\)=2-\(\dfrac{x}{2}\)
<=>\(\dfrac{8x+2x-8}{12}\)=\(\dfrac{24-6x}{12}\)
<=> 8x+2x-8=24-6x
<=>8x+2x+6x=24+8
<=>16x=32
<=>x=2
d) \(\dfrac{x-2}{x+2}\)-\(\dfrac{3}{x-2}\)= -\(\dfrac{2\left(x-11\right)}{4-x^2}\) ( ĐKXĐ: x\(\ne\)\(\pm\)2)
<=> \(\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{x^2-4}\)=\(\dfrac{2\left(x-11\right)}{x^2-4}\)
=> (x-2)2-3(x+2)=2(x-11)
<=> x2-4x+4-3x-6=2x-22
<=> x2-4x-3x-2x=-22-4+6
<=> x-9x+20=0
<=> (x-4)(x-5)=0
<=>\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) ( thỏa mãn diều kiện )
d) (x2+1)(x2-4x+4)=0
=> x2-4x+4=0 (x2+1\(\ge\)1 với mọi x)
=>(x-2)2 =0
=>x=2
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
a: =>-4x>16
=>x<-4
c: =>20x-25<=21-3x
=>23x<=46
=>x<=2
d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)
=>40x-100-90x+30<36-12x-30x+15
=>-50x-70<-42x+51
=>-8x<121
=>x>-121/8
1.
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\left(ĐKXĐ:x\ne1\right)\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\ \Leftrightarrow21x-9=2x-2\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\dfrac{7}{19}\left(TMĐK\right)\)
2.
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\left(ĐKXĐ:x\ne-\dfrac{2}{3};x\ne\dfrac{1}{3}\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\\ \Leftrightarrow-8x+1=-11x-14\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\left(TMĐK\right)\)
3.
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ \Leftrightarrow\left(\dfrac{1-x}{x+1}+3\right)\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}.\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{4+2x}{x+1}\left(x+1\right)=2x+3\\ \Leftrightarrow4+2x=2x+3\\ \Leftrightarrow4=3\)
Vô nghiệm.
Ta có :
\(VT=\left(\dfrac{1}{2}xy-\dfrac{1}{3}y\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy^2+\dfrac{1}{9}y^2\right)\)
\(=\dfrac{1}{8}x^3y^3+\dfrac{1}{12}x^2y^3+\dfrac{1}{18}xy^3-\dfrac{1}{12}x^2y^3-\dfrac{1}{18}xy^3-\dfrac{1}{27}y^3\)
\(=\dfrac{1}{8}x^3y^3-\dfrac{1}{27}y^3=VT\)
\(\Rightarrow dpcm\)
Vậy : ..............
\(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\left(ĐKXĐ:x\ne\pm1\right)\\ \Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Rightarrow x^2+x-2x=0\\ \Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\Rightarrow x=1\left(loại\right)\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={0}.
b)
\(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\)
quy đồng và khử mẫu phương trình trên, ta được:
\(\left(x+2\right)^2+3-2x=x^2+10\\ \Leftrightarrow x^2+4x+4-2x-x^2=10-3\)
\(\Leftrightarrow2x+4=7\Leftrightarrow2x=7-4=3\Rightarrow x=\dfrac{3}{2}\left(loại\right)\)
vậy phương trình đã cho vô nghiệm.
c)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\left(ĐKXĐ:x\ne\pm5\right)\)
\(\Leftrightarrow\dfrac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)^2}{\left(x+5\right)\left(x-5\right)}=\dfrac{20}{\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\left(x+5\right)^2-\left(x-5\right)^2=20\)
\(\Leftrightarrow x^2+25x+25-x^2+25x-25=20\\ \Leftrightarrow50x=20\Rightarrow x=\dfrac{2}{5}\)
vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{2}{5}\right\}\)
d)\(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\left(ĐKXĐ:x\ne\pm\dfrac{2}{3}\right)\)
quy đồng và khử mẫu phương trình trên, ta được:
\(\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\\ \Leftrightarrow9x^2+12x+4-18x+12-9x^2=0\\ \Leftrightarrow16-6x=0\Leftrightarrow6x=16\Rightarrow x=\dfrac{16}{6}\)
vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{16}{6}\right\}\)
e)\(\dfrac{3}{5x-1}+\dfrac{2}{3-5x}=\dfrac{4}{\left(1-5x\right)\left(5x-3\right)}\left(ĐKXĐ:x\ne\dfrac{1}{5};\dfrac{3}{5}\right)\)
quy đồng và khử mẫu phương trình trên, ta được:
\(3\left(3-5x\right)+2\left(5x-1\right)=4\\ \Leftrightarrow9-15x+10x-2=4\\ \Leftrightarrow-5x=-3\Rightarrow x=\dfrac{3}{5}\left(loại\right)\)
vậy phương trình đã cho vô nghiệm.
f)
\(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\left(ĐKXĐ:x\ne\pm\dfrac{1}{4}\right)\)
quy đồng và khử mẫu phương trình trên, ta được:
\(-3\left(4x+1\right)=2\left(4x-1\right)-8-6x\\ \Leftrightarrow-12x-3=8x-2-8-6x\\ \Leftrightarrow-14x=-7\Rightarrow x=\dfrac{1}{2}\)
vậy phương trình có tập nghiệm là \(S=\left\{\dfrac{1}{2}\right\}\)
g)
\(\dfrac{y-1}{y-2}-\dfrac{5}{y+2}=\dfrac{12}{y^2-4}+1\left(ĐKXĐ:y\ne\pm2\right)\)
quy đồng và khử mẫu phương trình trên, ta được:
\(\left(y-1\right)\left(y+2\right)-5\left(y-2\right)=12+y^2-4\\ \Leftrightarrow y^2+y-2-5y+10=12+y^2-4\\ \Leftrightarrow-4y+8=8\Leftrightarrow-4y=0\Rightarrow y=0\)
vậy phương trình có tập nghiệm là S={0}
h)
\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\left(ĐKXĐ:x\ne\pm1\right)\)
quy đồng và khử mẫu phương trình trên, ta được:
\(\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow x^2+2x+1-x^2+2x-1=4\\ \Leftrightarrow4x=4\Rightarrow x=1\)
vậy phương trình có tập nghiệm là S={1}.
i)
\(\dfrac{2x-3}{x+2}-\dfrac{x+2}{x-2}=\dfrac{2}{x^2-4}\left(ĐKXĐ:x\ne\pm2\right)\)
quy đồng và khử mẫu phương trình trên, ta được:
\(\left(2x-3\right)\left(x-2\right)-\left(x+2\right)=2\\ \Leftrightarrow2x^2-7x+6-x^2-4x-4=2\\ \Leftrightarrow x^2-11x=0\Rightarrow\left[{}\begin{matrix}x=0\\x-11=0\Rightarrow x=11\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={0;11}
j)
\(\dfrac{x-1}{x^2-4}=\dfrac{3}{2-x}\left(ĐKXĐ:x\ne\pm2\right)\)
quy đồng và khử mẫu phương trình trên, ta được:
\(x-1=-3\left(x+2\right)\Leftrightarrow x-1=-3x-6\\ \Leftrightarrow4x=5\Rightarrow x=\dfrac{5}{4}\)
vậy phương trình có tập nghiệm là \(S=\left\{\dfrac{5}{4}\right\}\)
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)