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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 :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)
\(\Leftrightarrow\frac{\left(x+2\right)^2}{8}-\frac{16\left(2x+1\right)}{8}=\frac{200}{8}+\frac{\left(x-2\right)^2}{8}\)
\(\Leftrightarrow\left(x+2\right)^2-32x-16=200+\left(x-2\right)^2\)
\(\Leftrightarrow x^2+4x+4-32x-16-200=x^2-4x+4\)
\(\Leftrightarrow x^2-28x-212-x^2+4x-4=0\)
\(\Leftrightarrow-24x=216\)
\(\Leftrightarrow x=-9\)
TL:
a)
\(\frac{\left(x+2\right)^2}{8}-\frac{16\left(2x+1\right)}{8}=\frac{200+\left(x-2\right)^2}{8}\)
\(\frac{x^2+4x+4-32x-16}{8}=\frac{200+x^2-4x+4}{8}\)
\(x^2-28x-12-200-x^2+4x-4=0\)
\(-24x-216=0\)
\(-24x=216\)
\(x=-9\)
Vậy x=-9
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}
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}
\(\Leftrightarrow\frac{8\left(x-2\right)^2}{24}-\frac{3\left(2x-3\right)\left(x+3\right)}{24}+\frac{4\left(x-4\right)^2}{24}=0\)
\(\Leftrightarrow8\left(x-2\right)^2-3\left(2x-3\right)\left(x+3\right)+4\left(x-4\right)^2=0\)
\(\Leftrightarrow8\left(x^2-4x+4\right)-3\left(2x^2+3x-9\right)+4\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow8x^2-32x+32-6x^2-9x+27+4x^2-32x+64=0\)
\(\Leftrightarrow6x^2-73x=-123\)
\(\Leftrightarrow x\left(6x-73\right)=-123\)
\(\Leftrightarrow x=-123\) hay \(6x=-50\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-123\\x=-\frac{50}{6}\end{matrix}\right.\)
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
a)\(\frac{1}{x-1}\)-\(\frac{3x2}{x3-1}\)=\(\frac{2x}{x2+x+1}\)
<=> \(\frac{1}{x-1}\)-\(\frac{3x2}{\left(x-1\right)\left(x2+x+1\right)}\)=\(\frac{2x}{x2+x+1}\) ĐKXĐ: x khác 1
<=> x2+x+1 - 3x2 = 2x(x-1)
<=>x2+x+1 - 3x2 = 2x2-2x
<=>x2-3x-1=0( đoạn này làm nhanh nhé)
<=>x2-2*\(\frac{3}{2}\)x +\(\frac{9}{4}\)-\(\frac{9}{4}\)-1=0
<=>(x-\(\frac{3}{2}\))2-\(\frac{13}{4}\)=0
<=>(x-\(\frac{3-\sqrt{13}}{2}\))(x-\(\frac{3+\sqrt{13}}{2}\))=0
\(\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}\)
b) pt... đkxđ x khác 1;2;3
<=> 3(x-3) +2(x-2)=x-1
<=> 3x-9 +2x-4 = x-1
<=> 4x= 12
<=> x=3 ( ko thỏa đk)
vậy pt vô nghiệm
\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\cdot\left(x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}\)
\(=\frac{x^2-4x+4}{3}-\frac{2x^2+3x-9}{8}+\frac{x^2-8x+16}{6}\)
\(=\frac{8x^2-32x+16}{24}-\frac{6x^2+9x-27}{24}+\frac{4x^2-32x+64}{24}\)
\(=\frac{8x^2-32x+16-6x^2-9x+27+4x^2-32x+64}{24}\)
\(=\frac{6x^2-73x+107}{24}\)