a, 3-4x(25-2x)=8x^2+x-30

<=> 3-100x+8x^2=8x^2+x-30

<=>3-100x+8x^2-8x^2-x+30=0

<=>-101x+33=0

<=>-101x=-33

<=>x=\(\dfrac{33}{101}\)

Vậy S={\(\dfrac{33}{101}\) }

b,(2x+1)(3x-2)=(5x-8)(2x+1)

<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)[(3x-2)-(5x-8)]=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(-2x+6)=0

=> 2x+1=0 hoặc -2x+6=0

+) 2x+1=0

<=>2x=-1

<=>x=-1/2

+)-2x+6=0

<=>-2x=-6

<=>x=3

vậy S={-1/2;3}

c,d, do mình lười quá nên mình ghi luôn kết quả nhé : c, x= \(\dfrac{1}{2}\)

d, x=5

\(\frac{3x+2}{x+4}+\frac{2x+1}{x-2}=5-\frac{x-32}{x^2+2x-8}\)

\(\Leftrightarrow\) \(\frac{\left(3x+2\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}+\frac{\left(2x+1\right)\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}=\frac{5\left(x+4\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}-\frac{x-32}{\left(x+4\right)\left(x-2\right)}\)

\(\Rightarrow\) (3x + 2)(x - 2) + (2x + 1)(x + 4) = 5(x + 4)(x - 2) - x + 32

\(\Leftrightarrow\) 3x² - 6x + 2x - 4 + 2x² + 8x + x + 4 = 5x² - 10x + 20x - 40 - x + 32

\(\Leftrightarrow\) 5x² + 5x = 5x² + 9x - 8

\(\Leftrightarrow\) 5x² + 5x - 5x² - 9x + 8 = 0

\(\Leftrightarrow\) -4x + 8 = 0

\(\Leftrightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy S = {2}

\(\frac{x+2m}{x+3}+\frac{x-m}{x-3}=\frac{mx\left(x+1\right)}{x^2-9}\) (đkxđ: x \(\ne\) \(\pm\) 3)

\(\Leftrightarrow\) \(\frac{\left(x+2m\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-m\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{mx\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}\)

\(\Rightarrow\) (x + 2m)(x - 3) + (x - m)(x + 3) = mx(x + 1)

\(\Leftrightarrow\) x² - 3x + 2mx - 6m + x² + 3x - mx - 3m - mx² - mx = 0

\(\Leftrightarrow\) (2 - m)x² - 9m = 0

Thay m = 1 ta được:

(2 - 1)x² - 9 . 1 = 0

\(\Leftrightarrow\) x² - 9 = 0

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

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(KTM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)

Vậy S = \(\varnothing\)

Thay m = 2 ta được:

(2 - 2)x² - 9 . 2 = 0

\(\Leftrightarrow\) -18 = 0

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!!

Lời giải:
Đặt $2^x=a; 3^{\frac{1}{x}}=b$. PT đã cho tương đương với:

\((2^x)^3+(3^{\frac{1}{x}})^3+2.2^x.3.3^{\frac{1}{x}}+2^x.3^2.3^{\frac{1}{x}}=125\)

\(\Leftrightarrow a^3+b^3+6ab+9ab=125\)

\(\Leftrightarrow a^3+b^3+15ab-125=0\)

\(\Leftrightarrow (a+b)^3-3ab(a+b)+15ab-5^3=0\)

\(\Leftrightarrow (a+b)^3-5^3-3ab(a+b-5)=0\)

\(\Leftrightarrow (a+b-5)[(a+b)^2+5(a+b)+25-3ab]=0\)

\(\Rightarrow \left[\begin{matrix} a+b-5=0\\ a^2+b^2+25-2ab+5a+5b=0\end{matrix}\right.\)

Nếu $a+b-5=0$

$\Leftrightarrow 2^x+3^{\frac{1}{x}}=5$

Hiển nhiên PT có nghiệm $x=1$. Còn 1 nghiệm nữa là nghiệm vô tỷ. Mình nghĩ với kiến thức lớp 9 mà không có thêm điều kiện ràng buộc của $x$ thì rất khó để giải.

Nếu $a^2+b^2+25-2ab+5a+5b=0$

$\Leftrightarrow \frac{(a-b)^2+(a+5)^2+(b+5)^2}{2}=0$

$\Rightarrow (a-b)^2=(a+5)^2=(b+5)^2=0$

$\Rightarrow a=b=-5$ (vô lý vì $2^x, 3^{\frac{1}{x}}$ luôn dương với mọi $x$)

@Nguyễn Việt Lâm bài pt này em giải mãi mak ch ra, nên anh giúp em nhé !!!

Nguyễn Việt Lâm

a) \(\frac{1-2x}{4}-2< \frac{1-5x}{8}+x\)

\(\Leftrightarrow\frac{2\left(1-2x\right)}{8}-\frac{16}{8}< \frac{1-5x}{8}+\frac{8x}{8}\)

\(\Leftrightarrow2-4x-16< 1-5x+8x\)

\(\Leftrightarrow-4x-14< 1-3x\)

\(\Leftrightarrow-x< 15\)

\(\Leftrightarrow x>-15\)

Vậy bất phương trình có tập nghiệm là: S ={x| x > -15}

b) \(\frac{1-x}{3}< \frac{x+4}{2}\)

\(\Leftrightarrow2\left(1-x\right)< 3\left(x+4\right)\)

\(\Leftrightarrow2-2x< 3x+12\)

\(\Leftrightarrow-5x< 10\)

\(\Leftrightarrow x>-2\)

Vậy bất phương trình có tập nghiệm là: S ={x| x > -2}

c) \(\frac{2x-3}{2}>\frac{8x-11}{6}\)

\(\Leftrightarrow3\left(2x-3\right)>8x-11\)

\(\Leftrightarrow6x-9>8x-11\)

\(\Leftrightarrow-2x>-2\)

\(\Leftrightarrow x< 1\)

Vậy bất phương trình có tập nghiệm là: S ={x| x < 1}

a) \(\frac{15x-10}{x^2+3}=0\)

<=> 15x - 10 = 0

<=> 5(3x - 2) = 0

<=> 3x - 2 = 0

<=> 3x = 2

<=> x = 2/3

b) ĐKXĐ: \(x\ne1;x\ne-3\)

<=>\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{x^2+2x-3}=0\)

<=> \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{\left(x-1\right)\left(x+3\right)}=0\)

<=> (3x - 1)(x + 3) - (2x + 5)(x - 1) - 8 = (x - 1)(x + 3)

<=> 3x² + 9x - x - 3 - 2x² + 2x - 5x + 5 - 8 = 0

<=> x² + 5x - 6 = 0

<=> (x - 1)(x + 6) = 0

<=> x - 1 = 0 hoặc x + 6 = 0

<=> x = 1 (ktm) hoặc x = -6 (tm)

=> x = -6

Bài 1:

ĐKXĐ: x≠1

Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)

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

\(\Leftrightarrow x^2+x+1+2x^2-5-4\left(x-1\right)=0\)

\(\Leftrightarrow x^2+x+1+2x^2-5-4x+4=0\)

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

Vì 3≠0

nên \(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Bài 2:

ĐKXĐ: x≠2; x≠3; \(x\ne\frac{1}{2}\)

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

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

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

\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(-x-4\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-12-x^2-2x+8=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(tm)

Vậy: x=-4

Bài 3:

ĐKXĐ: x≠1; x≠-1

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

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

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

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

\(\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3x\left(x^2-1\right)+3x\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1-3x^3+3x+3x^3-6x^2+3x=0\)

\(\Leftrightarrow-6x^2+10x=0\)

\(\Leftrightarrow2x\left(-3x+5\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{5}{3}\right\}\)

Bài 4:

ĐKXĐ: x≠1; x≠-3

Ta có: \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)

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

\(\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0\)

\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)

\(\Leftrightarrow13x-1=0\)

\(\Leftrightarrow13x=1\)

hay \(x=\frac{1}{13}\)(tm)

Vậy: \(x=\frac{1}{13}\)

Bài 5:

ĐKXĐ: x≠1; x≠-2

Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)

\(\Leftrightarrow\frac{x+2}{\left(x-1\right)\left(x+2\right)}-\frac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{3}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Leftrightarrow x+2-7\left(x-1\right)-3=0\)

\(\Leftrightarrow x+2-7x+7-3=0\)

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

\(\Leftrightarrow-6\left(x-1\right)=0\)

Vì -6≠0

nên x-1=0

hay x=1(ktm)

Vậy: x∈∅

Bài 6:

ĐKXĐ: x≠4; x≠2

Ta có: \(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{-\left(x^2-6x+8\right)}=0\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Bài 7:

ĐKXĐ: x≠1; x≠-2; x≠-1

Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{7}{x+2}+\frac{3}{x^2-1}=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}-\frac{7\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)\left(x+1\right)}+\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2+3x+2-7\left(x^2-1\right)+3x+6=0\)

\(\Leftrightarrow x^2+3x+2-7x^2+7x+3x+6=0\)

\(\Leftrightarrow-6x^2+13x+8=0\)
\(\Leftrightarrow-6x^2+16x-3x+8=0\)

\(\Leftrightarrow2x\left(-3x+8\right)+\left(-3x+8\right)=0\)

\(\Leftrightarrow\left(-3x+8\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+8=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{-1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};\frac{-1}{2}\right\}\)

\( 1)\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\ DK:x \ne 1\\ \Leftrightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^3} - 1}}\\ \Leftrightarrow {x^2} + x + 1 + 2{x^2} - 5 = 4x - 4\\ \Leftrightarrow 3{x^2} - 3x = 0\\ \Leftrightarrow 3x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\left( {tm} \right)\\ x = 1\left( {ktm} \right) \end{array} \right.\\ 2)\dfrac{{x + 4}}{{2{x^2} - 5x + 2}} + \dfrac{{x + 1}}{{2{x^2} - 7x + 3}} = \dfrac{{2x + 5}}{{2{x^2} - 7x + 3}}\\ + DK:x \ne \dfrac{1}{2};x \ne 2;x \ne 3\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {2x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}} = \dfrac{{2x + 5}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}}\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x - 2} \right) = \left( {2x + 5} \right)\left( {x - 2} \right)\\ \Leftrightarrow {x^2} + x - 12 + {x^2} - x - 2 = 2{x^2} + x - 10\\ \Leftrightarrow x = - 4\left( {tm} \right)\\ 3)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = 3x\left( {1 - \dfrac{{x - 1}}{{x + 1}}} \right)\\ DK:x \ne \pm 1\\ \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} = 3x\left( {x - 1} \right)\left( {x + 1 - x + 1} \right)\\ \Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 6x\left( {x - 1} \right)\\ \Leftrightarrow 4x = 6{x^2} - 6x\\ \Leftrightarrow 2x\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{5}{3} \end{array} \right.\left( {tm} \right) \)

Còn lại tương tự mà làm nhé!

Câu 1 :

a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)

=> \(3x-3-2x-6=-15\)

=> \(3x-3-2x-6+15=0\)

=> \(x=-6\)

Vậy phương trình có nghiệm là x = -6 .

b, Ta có : \(3\left(x-1\right)+2=3x-1\)

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

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

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)

=> \(14-35x-5=16-24x\)

=> \(14-35x-5-16+24x=0\)

=> \(-35x+24x=7\)

=> \(x=\frac{-7}{11}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .

Bài 2 :

a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)

=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)

=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)

=> \(x+15x-3=2x-16-10x-15\)

=> \(x+15x-3-2x+16+10x+15=0\)

=> \(24x+28=0\)

=> \(x=\frac{-28}{24}=\frac{-7}{6}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .

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

=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)

=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)

=> \(6x+24-30x+120=10x-15x+30\)

=> \(6x+24-30x+120-10x+15x-30=0\)

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

=> \(x=\frac{-114}{-19}=6\)

Vậy phương trình có nghiệm là x = 6 .

\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\)                              \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)

\(\Leftrightarrow\frac{\left(3\text{x}-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2\text{x}+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}\)

\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)

\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)

Vậy pt vô nghiệm