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KT
26 tháng 4 2018
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
12 tháng 4 2018
a) \(15x-3\left(3x-2\right)=45-5\left(2x-5\right)\)
\(\Leftrightarrow15x-9x+6=45-10x+25\)
\(\Leftrightarrow15x-9x+10x=45+25-6\)
\(\Leftrightarrow16x=64\)
\(\Leftrightarrow x=4\)
b) \(x^2-9+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\x+7=0\Leftrightarrow x=-7\end{matrix}\right.\)
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow x+4+x^2-4x+2x-8=5x-4\)
\(\Leftrightarrow x^2+x-4x+2x-5x=-4+8-4\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\Leftrightarrow x=6\end{matrix}\right.\)
KN
12 tháng 4 2018
a) 15x - 3(3x - 2) = 45 - 5(2x - 5)
\(\Leftrightarrow\) 15x - 9x + 6 = 45 - 10x + 25
\(\Leftrightarrow\) 6x + 10x = 70 - 6
\(\Leftrightarrow\) 16x = 64
\(\Leftrightarrow\) x = 4
Vậy.......................
b) x2 - 9 + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3) + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3 + 4) = 0
\(\Leftrightarrow\) (x - 3)(x + 7) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=3\end{matrix}\right.\)
Vậy........................
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\) (đk: x\(\ne\pm\)4)
\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}+\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5x-4}{\left(x+4\right)\left(x-4\right)}\)
\(\Leftrightarrow\) x + 4 + x2 - 4x + 2x - 8 = 5x - 4
\(\Leftrightarrow\) x2 - x - 5x - 4 + 4 = 0
\(\Leftrightarrow\) x2 - 6x = 0
\(\Leftrightarrow\) x(x - 6) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=6\left(tmđk\right)\end{matrix}\right.\)
Vậy...............
TN
27 tháng 2 2020
\(\frac{-5}{9}x+1=\frac{2}{3}x-10\)
\(\frac{-5}{9}x+\frac{9}{9}=\frac{6}{9}x-\frac{90}{9}\)
\(-5x+9=6x-90\)
\(-5x-6x=-90-9\)
\(-11x=-99\)
\(x=\frac{-99}{-11}=9\)
b. \(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)
\(\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)
\(\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)
\(\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)
x=30
Chúc bạn học tốt!!
27 tháng 4 2018
a đkxđ khi x khác 2 và -2 \(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2-\left(x-2\right)^2}{x^2-4}=\frac{4}{x^2-4}\)
\(\Rightarrow\left(x+2\right)^2-\left(x-2\right)^2=4\)\(\Rightarrow\left(x+2-x+2\right)\left(x+2+x-2\right)=4\Rightarrow4\cdot2x=4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)(thảo mãn)
27 tháng 4 2018
b đkxđ khi x+3 khác 0 suy ra x khác -3
\(\frac{x^2-9}{x+3}=\frac{\left(x-3\right)\left(x+3\right)}{x+3}=x-3=0\Rightarrow x=3\)(thảo mãn)
HQ
15 tháng 4 2018
a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1
b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)
\(\Leftrightarrow x< 5\)
c) \(\dfrac{-1}{2x+3}< 0\)
dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp
MD
26 tháng 5 2017
a) ĐKXĐ: \(x\ne2;4\)
\(\dfrac{x-3}{x-2}-\dfrac{x-2}{x-4}\) = \(\dfrac{16}{5}\)
<=> \(\dfrac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\) = \(\dfrac{16}{5}\)
<=> \(\dfrac{x^2-7x+12-x^2+4x-4}{\left(x-2\right)\left(x-4\right)}-\dfrac{16}{5}\) = 0
<=> \(\dfrac{5\left(-3x+8\right)}{5\left(x-2\right)\left(x-4\right)}-\dfrac{16\left(x^2-6x+8\right)}{5\left(x-2\right)\left(x-4\right)}\) = 0
=> \(-15x+40-16x^2+96x-128\) = 0
<=> \(-\left(16x^2-81x+88\right)\) = 0
<=> \(16x^2-81x+88\) = 0
<=> \(\left(16x^2-81x+\dfrac{6561}{64}\right)-\dfrac{929}{64}\) = 0
<=> \(\left(4x-\dfrac{81}{8}\right)^2\) = \(\dfrac{929}{64}\)
<=> \(\left[{}\begin{matrix}4x-\dfrac{81}{8}=\sqrt{\dfrac{929}{64}}\\4x-\dfrac{81}{8}=-\sqrt{\dfrac{929}{64}}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\dfrac{81+\sqrt{929}}{32}\\x=\dfrac{81-\sqrt{929}}{32}\end{matrix}\right.\)
Vậy .......................................... ( số xấu nhỉ!)
MD
26 tháng 5 2017
b) \(2x^2-6x+1\) = 0
<=> \(2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{7}{2}\) = 0
<=> \(2\left(x-\dfrac{3}{2}\right)^2\) = \(\dfrac{7}{2}\)
<=> \(\left(x-\dfrac{3}{2}\right)^2\) = \(\dfrac{7}{4}\)
<=> \(\left[{}\begin{matrix}x-\dfrac{3}{2}=\sqrt{\dfrac{7}{4}}\\x-\dfrac{3}{2}=-\sqrt{\dfrac{7}{4}}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\dfrac{3+\sqrt{7}}{2}\\x=\dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\)
Vậy .............................
c) \(3x^2+12x-66\) = 0
<=> \(3\left(x^2+4x+4\right)-78\) = 0
<=> \(3\left(x+2\right)^2\) = 78
<=> \(\left(x+2\right)^2\) = 26
<=> \(\left[{}\begin{matrix}x+2=\sqrt{26}\\x+2=-\sqrt{26}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-2+\sqrt{26}\\x=-2-\sqrt{26}\end{matrix}\right.\)
Vậy .................................
P/s: Yahoooooooooooooo.......xong rồi!
10 tháng 2 2018
a) \(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
\(3x^2+10x-8=5x^2-2x+10\)
\(3x^2-5x^2+10x+2x-8-10=0\)
\(-2x^2+12x-18=0\)
\(x^2-6x+9=0\)
\(\left(x-3\right)^2=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
b) \(\frac{x^2-x-6}{x-3}=0\)
\(\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-2x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-6=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{2}-\frac{5}{2}\right)\left(x-\frac{1}{2}+\frac{5}{2}\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
x5-36=0
x5=36
=>x\(\in\)rỗng
Ko cộng như vậy được đâu, Việt Anh ơi T_T!!