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\(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\)

=>\(x^2+x+1=4x\)

=>\(x^2-3x+1=0\)

\(\dfrac{x^5-3x^3-10x+12}{x^4+7x^2+15}\)

\(=\dfrac{x^5-3x^4+x^3+3x^4-9x^3+3x^2+5x^3-15x^2+5x+12x^2-36x+12+21x}{x^4+7x^2+15}\)

\(=\dfrac{x^3\left(x^2-3x+1\right)+3x^2\left(x^2-3x+1\right)+5x\left(x^2-3x+1\right)+12\left(x^2-3x+1\right)+21x}{x^4+7x^2+15}\)

\(=\dfrac{21x}{x^4-3x^3+x^2+3x^3-9x^2+3x+15x^2-45x+15+42x}\)

\(=\dfrac{21x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+15\left(x^2-3x+1\right)+42x}\)

\(=\dfrac{21x}{42x}=\dfrac{1}{2}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

23 tháng 2 2019

Câu 1:

Hỏi đáp Toán

23 tháng 2 2019

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 \(S=\left\{-1\right\}\)

6 tháng 2 2018

1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)

ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )

\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn

2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)

\(\Leftrightarrow-24x+144=-5x+30\)

\(\Leftrightarrow-24x+144+5x-30=0\)

\(\Leftrightarrow-19x+114=0\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: x=6

3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)

\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9-60-32x=0\)

\(\Leftrightarrow-2x-51=0\)

\(\Leftrightarrow-2x=51\)

hay \(x=-\dfrac{51}{2}\)

Vậy: \(x=-\dfrac{51}{2}\)

4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)

\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)

\(\Leftrightarrow2x+2-x+2=6x-3\)

\(\Leftrightarrow x+4-6x+3=0\)

\(\Leftrightarrow-5x+7=0\)

\(\Leftrightarrow-5x=-7\)

hay \(x=\dfrac{7}{5}\)

Vậy: \(x=\dfrac{7}{5}\)

21 tháng 1 2021

1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)

\(2\left(5x-2\right)=3\left(5-3x\right)\)

\(10x-4=15-9x\)

\(10x+9x=15+4\)

\(19x=19\)

\(x=1\)

Vậy \(x=1\)

28 tháng 1 2021

1/ ĐKXĐ : \(x\ne1\)

\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow19x=7\Leftrightarrow x=\dfrac{7}{19}\left(tm\right)\)

Vậy...

b/ \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) ĐKXĐ : \(x\ne-1\)

\(\Leftrightarrow12-28x=1+x\)

\(\Leftrightarrow11=29x\Leftrightarrow x=\dfrac{11}{29}\) \(\left(tm\right)\)

Vậy....

c/ ĐKXĐ : \(x\ne0\)

\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2x^2-12=2x^2+3x\)

\(\Leftrightarrow3x=-12\Leftrightarrow x=-4\) \(\left(tm\right)\)

Vậy...

4/ ĐKXĐ : \(x\ne-\dfrac{2}{3}\)

\(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)

\(\Leftrightarrow6x^2+4x-3x-2=5\)

\(\Leftrightarrow6x^2+x-7=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)

Vậy....

5,6 Tương tự nhé !

 

 

1)ĐKXĐ: \(x\ne1\)

Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)

\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow21x-9-2x+2=0\)

\(\Leftrightarrow19x-7=0\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\dfrac{7}{19}\)(nhận)

Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)

2) ĐKXĐ: \(x\ne-1\)

Ta có: \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)

\(\Leftrightarrow4\left(3-7x\right)=x+1\)

\(\Leftrightarrow12-28x-x-1=0\)

\(\Leftrightarrow-29x+11=0\)

\(\Leftrightarrow-29x=-11\)

\(\Leftrightarrow x=\dfrac{11}{29}\)

Vậy: \(S=\left\{\dfrac{11}{29}\right\}\)

3) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)

\(\Leftrightarrow2x^2-12=2x^2+6x\)

\(\Leftrightarrow2x^2-12-2x^2-6x=0\)

\(\Leftrightarrow-6x-12=0\)

\(\Leftrightarrow-6x=12\)

\(\Leftrightarrow x=-2\)

Vậy: S={-2}

14 tháng 6 2017

1

\(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)

=> \(-3x^2+15x+5x-5+3x^2=4-x\)

=> \(20x-5=4-x\)

=> \(21x=9\)

=> \(x=\dfrac{3}{7}\)

Vậy x = \(\dfrac{3}{7}\)

2,

\(7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)

=> \(7x^2-14x-5x+5=7x^2+3\)

=> \(-14x-5x+5=3\)

=> \(-19x=-2\)

=> \(x=\dfrac{2}{19}\)

Vậy \(x=\dfrac{2}{19}\)

3,

\(3\left(5x-1\right)-x\left(x-2\right)+x^2-13x=7\)

=> \(15x-3-x^2+2x+x^2-13x=7\)

=> \(4x-3=7\)

=> 4x = 10

=> x = \(\dfrac{5}{2}\)

Vậy x = \(\dfrac{5}{2}\)

4,

\(\dfrac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)=12\)

=> \(2x^2-3x-2x^2+10x=12\)

=> 7x = 12

=> x = \(\dfrac{12}{7}\)

Vậy x = \(\dfrac{12}{7}\)

14 tháng 6 2017

undefinedundefinedundefinedundefinedKkk

a) ĐKXĐ: \(x\notin\left\{-1;-2;2\right\}\)

Ta có: \(\dfrac{1}{x^2+3x+2}-\dfrac{3}{x^2-x-2}=\dfrac{-1}{x^2-4}\)

\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}-\dfrac{3}{\left(x-2\right)\left(x+1\right)}=\dfrac{-1}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{x-2}{\left(x+1\right)\left(x+2\right)\left(x-2\right)}-\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-1\left(x+1\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x-2-3x-6=-x-1\)

\(\Leftrightarrow-2x-8+x+1=0\)

\(\Leftrightarrow-x-7=0\)

\(\Leftrightarrow-x=7\)

hay x=-7(thỏa ĐK)

Vậy: S={-7}

8 tháng 3 2021

a) ĐKXĐ: x∉{−1;−2;2}x∉{−1;−2;2}

Ta có: 1x2+3x+2−3x2−x−2=−1x2−41x2+3x+2−3x2−x−2=−1x2−4

⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)

⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)

Suy ra: x−2−3x−6=−x−1x−2−3x−6=−x−1

⇔−2x−8+x+1=0⇔−2x−8+x+1=0

⇔−x−7=0⇔−x−7=0

⇔−x=7⇔−x=7

hay x=-7(thỏa ĐK)

Vậy: S={-7}

Đọc tiếp

a) ĐKXĐ: x∉{−1;−2;2}x∉{−1;−2;2}

Ta có: 1x2+3x+2−3x2−x−2=−1x2−41x2+3x+2−3x2−x−2=−1x2−4

⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)

⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)

Suy ra: x−2−3x−6=−x−1x−2−3x−6=−x−1

⇔−2x−8+x+1=0⇔−2x−8+x+1=0

⇔−x−7=0⇔−x−7=0

⇔−x=7⇔−x=7

hay x=-7(thỏa ĐK)

Vậy: S={-7}

Đọc tiếp

f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

=>28-6(x+2)=-9-5(x-4)

=>28-6x-12=-9-5x+20

=>-6x+16=-5x+11

=>-x=-5

=>x=5

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11 tháng 3 2023

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