d) \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x-2}{5}-5\)

\(\Leftrightarrow\frac{5\left(5x+2\right)}{30}-\frac{10\left(8x-1\right)}{30}=\frac{6\left(4x-2\right)}{30}-\frac{150}{30}\)

\(\Leftrightarrow25x+10-80x+10=24x-12-150\)

\(\Leftrightarrow25x-80x-24x=-12-150-10-10\)

\(\Leftrightarrow-79x=-182\)

\(\Leftrightarrow x=\frac{182}{79}\).

Vậy tập nghiệm phương trình \(s=\left\{\frac{182}{79}\right\}\)

a)\(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\)

\(\Leftrightarrow\frac{3\left(3x+2\right)}{6}-\frac{3x+1}{6}=\frac{10}{6}+\frac{12x}{6}\)

\(\Leftrightarrow9x+6-3x+1=10+12x\)

\(\Leftrightarrow9x-3x-12x=10-6-1\)

\(\Leftrightarrow-6x=3\)

\(\Leftrightarrow x=\frac{-1}{2}\).

Vậy tập nghiệm phương trình \(S=\left\{\frac{-1}{2}\right\}\)

A B C D d M N I

a, Xét tam giác ADC có : MN // DC hay MI // DC 

Theo định lí Ta - lét ta có : \(\frac{MA}{MD}=\frac{IA}{IC}\)

b, Xét tam giác ABC có : AB // MN hay AB // IN 

Theo định lí Ta - lét ta có : \(\frac{BN}{NC}=\frac{IA}{IC}\)

mà \(\frac{MA}{MD}=\frac{IA}{IC}\)( cmt )

Suy ra : \(\frac{MA}{MD}=\frac{NB}{NC}\)

a. \(\frac{5x-2}{3}=\frac{5x-3x}{2}\)
\(\Leftrightarrow2.\left(5x-2\right)=3.\left(5x-3x\right) \)
\(\Leftrightarrow10x-4=15x-9x\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy...
b. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\left(1\right)\)
MC = 36.
pt (1) <=>
\(\frac{3\left(10x+3\right)}{36}=\frac{36}{36}+\frac{4\left(6+8x\right)}{36}\)
=> 3.(10x+3) = 36 + 4(6+8x)
<=> 30x+9 = 36+24+32x
<=> -2x = 51
<=> x = \(\frac{-51}{2}\)
Vậy...
c. \(\frac{7x-1}{6}+2=\frac{16-x}{5}\left(2\right)\)
MC = 30.
pt (2) <=>
\(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)
=> 5(7x-1) + 60x = 6(16-x)
<=> 35x-5 + 60x = 96-6x
<=> 101x = 101
<=> x = 1
Vậy...
d. \(\frac{3x+2}{2}-\frac{3x+1}{6}=5\) (3)
MC = 12.
pt (3)<=>
\(\frac{6\left(3x+2\right)}{12}-\frac{2\left(3x+1\right)}{12}=\frac{60}{12}\)
=> 6(3x+2) - 2(3x+1) = 60
<=> 18x+12 - 6x-2 = 60
<=> 12x = 50
<=> x = \(\frac{25}{6}\)
Vậy...
e. \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\) (4)
MC = 30.
pt (4) <=>
\(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> 6(x+4) - 30x + 120 = 10x - 15(x-2)
<=> 6x+24 - 30x + 120 = 10x - 15x+30
<=> -19x = -114
<=> x = \(\frac{114}{19}=6\)
Vậy...

ai làm ơn giúp tớ đi mà TwT

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

ĐKXĐ: \(x\ne\pm2\)

\(\Leftrightarrow \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) - \left( {x - 1} \right)\left( {x - 2} \right)}}{{{x^2} - 4}} = \dfrac{{2\left( {{x^2} + 2} \right)}}{{{x^2} - 4}}\\ \Leftrightarrow {x^2} + 3x + 2 - \left( {{x^2} - 3x + 2} \right) = 2{x^2} + 4\\ \Leftrightarrow 6x = 2{x^2} + 4\\ \Leftrightarrow - 2{x^2} + 6x - 4 = 0\\ \Leftrightarrow 2{x^2} - 6x + 4 = 0\\ \Leftrightarrow {x^2} - 3x + 2 = 0\\ \Leftrightarrow {x^2} - 2x - x + 2 = 0\\ \Leftrightarrow x\left( {x - 2} \right) - \left( {x - 2} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\ x - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\left( {KTM} \right)\\ x = 1\left( {TM} \right) \end{array} \right. \)

Vậy \(x=1\)

\(b)\dfrac{{x - 1}}{{x + 2}} - \dfrac{x}{{x - 2}} = \dfrac{{5x - 2}}{{4 - {x^2}}} \)

ĐKXĐ: \(x\ne\pm2\)

\( \Leftrightarrow \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right) - x\left( {x + 2} \right)}}{{{x^2} - 4}} = \dfrac{{2 - 5x}}{{{x^2} - 4}}\\ \Leftrightarrow {x^2} - 3x + 2 - {x^2} - 2x = 2 - 5x\\ \Leftrightarrow 0x = 0\left( {VSN} \right) \)

Vậy phương trình vô số nghiệm

\(c)\dfrac{{x - 2}}{{2 + x}} - \dfrac{3}{{x - 2}} = \dfrac{{2\left( {x - 11} \right)}}{{{x^2} - 4}}\)

ĐKXĐ: \(x\ne\pm2\)

\( \Leftrightarrow \dfrac{{\left( {x - 2} \right)\left( {x - 2} \right) - 3\left( {x + 2} \right)}}{{{x^2} - 4}} = \dfrac{{2x - 22}}{{{x^2} - 4}}\\ \Leftrightarrow {x^2} - 4x + 4 - 3x - 6 = 2x - 22\\ \Leftrightarrow {x^2} - 9x + 20 = 0\\ \Leftrightarrow {x^2} - 4x - 5x + 20 = 0\\ \Leftrightarrow x\left( {x - 4} \right) - 5\left( {x - 4} \right) = 0\\ \Leftrightarrow \left( {x - 4} \right)\left( {x - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 4 = 0\\ x - 5 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 4\left( {TM} \right)\\ x = 5\left( {TM} \right) \end{array} \right. \)

Vậy \(x=4,x=5\)

\(\frac{x-2}{71}+\frac{x-4}{69}=\frac{x-6}{67}+\frac{x-8}{65}\)

\(\Leftrightarrow\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}=\frac{x-73}{67}+\frac{x-73}{65}\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}-\frac{x-73}{67}-\frac{x-73}{65}=0\)

\(\Leftrightarrow\left(x-73\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)

Mà \(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\ne0\)

\(x-73=0\Leftrightarrow x=73\)

\(\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}=\frac{x-73}{67}+\frac{x-73}{65}\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)

\(\Rightarrow...\)

Well, it's ez, right? Hướng dẫn thôi nhé :> (*gớm, xài brain nhiều vào :V*)

a, ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

\(\frac{x}{2x+2}-\frac{2x}{x^2-2x-3}=\frac{x}{6-2x}\\ \Leftrightarrow\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=\frac{x}{-2\left(x-3\right)}\\ \Leftrightarrow\frac{x\left(x-3\right)-4x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{-x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}\Leftrightarrow...\)

Đến đây khử mẫu, giải PT và xét nghiệm với ĐKXĐ nhé (cứ thấy linh tinh với ĐKXĐ là cho outplay lun :>)

b, ĐKXĐ: \(x\notin\left\{2;3\right\}\)

\(\frac{5}{-x^2+5x-6}+\frac{x+3}{2-x}=0\\ \Leftrightarrow\frac{-5}{-\left(x-2\right)\left(x-3\right)}+\frac{x+3}{2-x}=0\\\Leftrightarrow\frac{-5}{\left(2-x\right)\left(x-3\right)}=\frac{-\left(x+3\right)\left(x-3\right)}{\left(2-x\right)\left(x-3\right)}\Leftrightarrow...\)

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

\(\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-4}{x+2}\\ \Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x-1}=\frac{-4}{x+2}\\ \Leftrightarrow\frac{3-\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}=\frac{-4\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}\Leftrightarrow...\)

Thế thui, chúc bạn học tốt nha.

Ahuhu, không ai biết cách giải ư ? T^T

Bài giải

x³ - 2x - 4

= ( x³ - 2x² ) + ( 2x² - 4x ) + ( 2x - 4 )

= x²( x - 2 ) + 2x( x - 2 ) + 2( x - 2 )

= ( x - 2 )( x² + 2x + 2 )

Vậy x³ - 2x - 4 = ( x - 2 )( x² + 2x + 2 )

Ta có: x^3-2x-4

= x^3 -8 -2x+4

= (x-2)(x^2+2x+4)-2(x-2)

= (x-2)(x^2+2x+2)

Ta đặt: a/b = a/d =k

  => a = b.k, c=d.k

 Ta có: a² + a.c/c² - a.c=b² + b.d/d² - b.d

 Vế trái:  => (b.k)² + (b.k)(d.k)/(d.k)² - (b.k)(d.k)

  => b².k² + k(b.d)/d².k² - k.(b.d)

 Ta lược bỏ các chữ giống nhau, ta được:

  => b²/d²

 Vế phải: b² +b.d/d² - b.d

 Ta cũng lược bỏ những chữa giống nhau ta được:

  => b²/d² 

Vậy a2 +a.c/c2 + a.c = b2 + b.d/d2 - b.d