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10 tháng 9 2023

\(...\Rightarrow2x-\dfrac{1}{27}=8\)

\(\Rightarrow2x=8+\dfrac{1}{27}\)

\(\Rightarrow2x=\dfrac{217}{27}\)

\(\Rightarrow x=\dfrac{217}{54}\)

27 tháng 8 2021

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

27 tháng 8 2021

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

\(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x2 + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x2 = 48 - 312 - 520 + 312

⇔ 1632x - 288x2 = -472

⇔ -288x2 + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

29 tháng 1 2020

                                                                    Bài giải

a, \(-2\left(2x-8\right)+3\left(4-2x\right)=-72-5\left(3x-7\right)\)

\(-4x+8+12-6x=-72-15x+7\)

\(-10x+20=-65-15x\)

\(-10x+15x=-65-20\)

\(5x=-85\)

\(x=-85\text{ : }5\)

\(x=-17\)

b, \(3\left|2x^2-7\right|=33\)

\(\left|2x^2-7\right|=33\text{ : }3\)

\(\left|2x^2-7\right|=11\)

\(\Rightarrow\orbr{\begin{cases}2x^2-7=-11\\2x^2-7=11\end{cases}}\Rightarrow\orbr{\begin{cases}2x^2=-4\text{ ( loại ) }\\2x^2=18\end{cases}}\Rightarrow\text{ }x^2=9\text{ }\Rightarrow\text{ }x=\pm3\)

\(\Rightarrow\text{ }x=\pm3\)

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

a: \(=\dfrac{4x-8+2x+4-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}=\dfrac{6}{x+2}\)

b: \(=\dfrac{-x+7x-4}{3x-2}=\dfrac{6x-4}{3x-2}=2\)

c: \(=\dfrac{x}{2x+1}-\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}-\dfrac{\left(x-2\right)}{2x-1}\)

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

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

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

d: \(=\dfrac{5}{2x-3}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-99}\)

\(=\dfrac{10x+15+4x-6+2x-33}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x-24}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{8}{2x+3}\)

31 tháng 3 2020

a)

\(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\\ \Leftrightarrow\frac{6x+30}{24}-\frac{16x-24}{24}-\frac{18x-3}{24}-\frac{4x-2}{24}=0\\ \Leftrightarrow\frac{6x+30-16x+24-18x+3-4x+2}{24}=0\\ \Leftrightarrow\frac{59-32x}{24}=0\\ \Rightarrow59-32x=0\\ \Rightarrow x=\frac{59}{32}\)

b)

\(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\\ \Leftrightarrow\frac{6x+24-30x+120-10x+15x-30}{30}=0\\ \Leftrightarrow\frac{114-19x}{30}=0\\ \Rightarrow114-19x=0\\ \Rightarrow x=\frac{-144}{-19}=6\\ \Rightarrow x=6\)

c)

\(x^2-3x+2=0\\ \Leftrightarrow2-x-2x+x^2=0\\ \Leftrightarrow2\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

8 tháng 2 2020

dễ mà

8 tháng 2 2020

a) -2(2x - 8) + 3(4 - 2x) = -72  - 5(3x - 7)

=> -4x + 18 + 12 - 6x = -72 - 15x + 35

=> -10x + 15x = -37 - 30

=> 5x = -37

=> x = -7,4

b) 3|2x2 - 7| = 33

=> |2x2 - 7| = 11

=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}}\)

=> \(\orbr{\begin{cases}2x^2=18\\2x^2=-4\left(loại\right)\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

6 tháng 7 2020

Trả lời 

a, \(3x=-75\)

      \(x=-25\)     

\(\frac{x}{7}=\frac{-6}{8}\)

    \(8x=-42\)

      \(x=\frac{-21}{4}\)

c, \(\frac{-5}{6}+x=\frac{7}{12}-\frac{1}{3}\)

      \(\frac{-5}{6}+x=\frac{1}{4}\)

                     \(x=\frac{13}{12}\)

d, \(\frac{1}{2}.x-\frac{33}{55}=\frac{20}{50}\)

      \(\frac{1}{2}.x=1\)

        \(x=2\)

Học tốt 

6 tháng 7 2020

a,  3X = -75

       X = -75 :3

       X = -25

b,  X x 7 = - 6 x 8

    X x 7 = -48

   X       = -48 :7

   X       = -48/7