Giiari phương trình:
\(\frac{\left(7x+1\right)\left(x-2\right)}{10}+\frac{2}{5}=\frac{\left(x-2\right)^2}{5}+\frac{\left(x-1\right)\left(x-3\right)}{2}\)
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11 tháng 3 2020
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
11 tháng 3 2020
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}
8 tháng 1 2020
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
27 tháng 5 2020
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne-2;x\ne-3\)
\(\Leftrightarrow x+3+x+2=1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\) (không nhận)
Vậy : \(S=\varnothing\)
27 tháng 5 2020
Giai phương trình sau :
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
ĐKXĐ : \(x\ne1;x\ne-5\)
Với điều kiện trên ta có :
\(\Leftrightarrow\)\(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{-3}{x-1}=\frac{5}{x+5}\)
\(\Leftrightarrow10-3\left(x+5\right)=5\left(x-1\right)\)
\(\Leftrightarrow10-3x-15=5x-5\)
\(\Leftrightarrow-8x=0\)
\(\Leftrightarrow x=0\) (nhận)
Vậy : \(S=\left\{0\right\}\)
28 tháng 5 2020
a,
\(\frac{10}{(x+5)(x-1)}+\frac{3}{x-1}=\frac{5}{x+5}\)
\(\frac{10}{(x+5)(x-1)}+\frac{3(x+5)}{(x+5)(x-1)}=\frac{5(x-1)}{(x+5)(x-1)}\)
\(\Leftrightarrow\)10+3x+15=5x-5
\(\Leftrightarrow\)25+3x=5x-5
\(\Leftrightarrow\)25+3x-5x+5=0
\(\Leftrightarrow\)30-2x=0
\(\Leftrightarrow\)-2x=-30
\(\Leftrightarrow\)x=15
vậy pt có ngiệm là x=15
BH
2 tháng 3 2019
a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)
b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)
c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)
\(\frac{\left(7x+1\right)\left(x-2\right)}{10}+\frac{2}{5}=\frac{\left(x-2\right)^2}{5}+\frac{\left(x-1\right)\left(x-3\right)}{2}\)
⇔ \(\frac{7x^2-13x-2}{10}+\frac{2}{5}-\frac{x^2-4x+4}{5}-\frac{x^2-4x+3}{2}=0\)
⇔ \(\frac{7x^2-13x-2}{10}+\frac{4}{10}-\frac{2\left(x^2-4x+4\right)}{10}-\frac{5\left(x^2-4x+3\right)}{10}=0\)
⇔ \(\frac{7x^2-13x-2}{10}+\frac{4}{10}-\frac{2x^2-8x+8}{10}-\frac{5x^2-20x+15}{10}=0\)
⇔ \(\frac{7x^2-13x-2+4-2x^2+8x-8-5x^2+20x-15}{10}=0\)
⇔ \(\frac{15x-21}{10}=0\)
⇔ 15x - 21 = 0
⇔ x = 21/15
\(\frac{\left(7x+1\right)\left(x-2\right)}{10}+\frac{2}{5}=\frac{\left(x-2\right)^2}{5}+\frac{\left(x-1\right)\left(x-3\right)}{2}\)
\(\Leftrightarrow\frac{7x^2-13x-2}{10}+\frac{4}{10}=\frac{2\left(x^2-4x+4\right)}{10}+\frac{5\left(x^2-4x+3\right)}{10}\)
\(\Leftrightarrow\frac{7x^2-13x-2+4}{10}=\frac{2x^2-8x+8+5x^2-20x+15}{10}\)
\(\Leftrightarrow\frac{7x^2-13x+2}{10}=\frac{7x^2-28x+23}{10}\)
\(\Leftrightarrow\frac{7x^2-13x+2}{10}-\frac{7x^2-28x+23}{10}=0\)
\(\Leftrightarrow7x^2-13x+2-7x^2+28x-23=0\)
\(\Leftrightarrow15x-21=0\)
\(\Leftrightarrow15x=21\)
\(\Leftrightarrow x=\frac{21}{15}=\frac{7}{5}\)