a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

=>\(\dfrac{3}{2}\sqrt{x-3}=3\)

=>\(\sqrt{x-3}=2\)

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

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

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

a) 2x - 3 > 3(x - 2)

⇔ 2x - 3 > 3x - 6

⇔ 2x - 3x > -6 + 3

⇔ -x > -3

⇔ x < 3

Vậy S = {x | x < 3}

b) (12x + 1)/12 ≤ (9x + 1)/3 - (8x + 1)/4

⇔ 12x + 1 ≤ 4(9x + 1) - 3(8x + 1)

⇔ 12x + 1 ≤ 36x + 4 - 24x - 3

⇔ 12x - 36x + 24x ≤ 4 - 3 - 1

⇔ 0x ≤ 0 (luôn đúng với mọi x)

Vậy S = R

a: =>2x-3>3x-6

=>-x>-3

=>x<3

b: =>12x+1<=36x+4-24x-3

=>12x+1<=12x+1

=>0x<=0(luôn đúng)

a) \(2x-3>3\left(x-2\right)\)

\(\Leftrightarrow2x-3>3x-6\)

\(\Leftrightarrow2x-3x>-6+3\)

\(\Leftrightarrow-x>-3\)

\(\Leftrightarrow x< 3\)

Vậy bất phương trình có nghiệm là \(x< 3\)

b) \(\dfrac{12x+1}{12}\le\dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow\dfrac{12x+1}{12}\le\dfrac{\left(9x+1\right).4}{3.4}-\dfrac{\left(8x+1\right)3}{4.3}\)

\(\Leftrightarrow12x+1\le36x+4-24x-3\)

\(\Leftrightarrow12x-36x+24x\le4-3-1\)

\(\Leftrightarrow0x\le0\)

Vậy bất phương trình vô nghiệm

Ở câu (h) mình quên gạch chân phân số, bạn thông cảm nha <3

g: =>12x+1>=36x+12-24x-3

=>12x+1>=12x+9(loại)

h: =>6(x-1)+4(2-x)<=3(3x-3)

=>6x-6+8-4x<=9x-9

=>2x+2<=9x-9

=>-7x<=-11

=>x>=11/7

i: =>4x^2-12x+9>4x^2-3x

=>-12x+9>-3x

=>-9x>-9

=>x<1

$ĐKXĐ:x \neq -4;-5;-6;-7$

$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$

$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$

$⇔x^2+11x+28=54$

$⇔x^2+11x-26=0$

$⇔x^2-2x+13x-26=0$

$⇔(x-2)(x+13)=0$

$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)

Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$

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

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

giúp mình với ạ câu nào cũng được

\(ĐK:x\ne\pm1\)

\(\dfrac{5x+3}{x-1}+\dfrac{3x}{x+1}=\dfrac{9x-4}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow\dfrac{\left(5x+3\right)\left(x+1\right)+3x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{9x-4}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow\left(5x+3\right)\left(x+1\right)+3x\left(x-1\right)=9x-4\)

\(\Leftrightarrow5x^2+5x+3x+3+3x^2-3x-9x+4=0\)

\(\Leftrightarrow8x^2-4x+7=0\)

Vậy pt vô nghiệm

\(\Leftrightarrow\left(5x+3\right)\left(x+1\right)+3x\left(x-1\right)=9x-4\)

\(\Leftrightarrow5x^2+5x+3x+3+3x^2-3x-9x+4=0\)

\(\Leftrightarrow8x^2-4x+7=0\)

\(\text{Δ}=\left(-4\right)^2-4\cdot8\cdot7=-208< 0\)

Do đó: Phương trình vô nghiệm