\(\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{4}{3}+\dfrac{19}{3}=\dfrac{1+2+4+19}{3}=\dfrac{26}{3}\)

\(\dfrac{3}{4}+\dfrac{4}{4}+\dfrac{5}{4}+\dfrac{6}{4}+\dfrac{x}{4}+\dfrac{8}{2}+\dfrac{9}{4}\)

=\(\dfrac{3}{4}+\dfrac{4}{4}+\dfrac{5}{4}+\dfrac{6}{4}+\dfrac{x}{4}+\dfrac{16}{4}+\dfrac{9}{4}\)

=\(\dfrac{3+4+5+6+x+16+9}{4}=\dfrac{43+x}{4}\)

Cảm ơn và chúc Lê Minh Quang học tốt nhé!

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1/3+2/3+4/3+19/3=26/3
3/4+4/4+5/4+6/4+x/2+8/2+9/4=43+x/4

a. \(\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{-33}{25}\)

\(\Rightarrow\dfrac{11}{10}x=\dfrac{-33}{25}\)

\(\Rightarrow x=\dfrac{-33}{25}:\dfrac{11}{10}=\dfrac{-6}{5}\)

Vậy.........

b. \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}+\dfrac{-3}{7}:x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{4}{9}=0\\\dfrac{1}{2}+\dfrac{-3}{7}:x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{-3}{7}:x=\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{6}{7}\end{matrix}\right.\)

Vậy................

a, 1/2xX+3/5xX=-33/25

Xx(1/2+3/5)=-33/25

Xx11/10=-33/25

X=-6/5

b,

ĐK: `x \ne \pm 3`

`(2(9+2x))/(x^2-9)=2/(x-3)-1/(x+3)`

`<=>2(2x+9)=2(x+3)-(x-3)`

`<=>4x+18=2x+6-x+3`

`<=>4x+18=x+9`

`<=>3x=-9`

`x=-3 (L)`

Vậy `S=∅`.

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

\(\Leftrightarrow\dfrac{18+4x}{\left(x-3\right)\left(x+3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow18+4x-2x-6+x-3=0\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3\left(x+3\right)=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

\(S=\left\{-3\right\}\)

ĐKXĐ:       x ≠ 3  ; x ≠  -3

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

⇒  \(2\left(9+2x\right)=2\left(x+3\right)-1\left(x-3\right)\)

Ta có: \(2\left(9+2x\right)=2\left(x+3\right)-1\left(x-3\right)\)

              \(18+4x=2x+6-x+3\)

       \(4x-2x+x=6+3-18\)

                       \(3x=0\)

                 \(\Rightarrow x\) vô nghiệm

Vậy phương trình vô nghiệm

3/7x2=6/7

5/12x1=5/12

3x4/7=12/7

0x5/9=0

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

\(B=\dfrac{x^2+39}{\left(x-3\right)\left(x+3\right)}+\dfrac{8}{x+3}-\dfrac{1}{x-3}\)

\(=\dfrac{x^2+39+8x-24-x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+7x+12}{\left(x-3\right)\left(x+3\right)}\)

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

\(B=\dfrac{x^2+39}{x^2-9}+\dfrac{8}{x+3}+\dfrac{1}{3-x}\) (ĐK: \(x\ne3,x\ne-3\))

\(B=\dfrac{x^2+39}{\left(x+3\right)\left(x-3\right)}+\dfrac{8\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{1}{x-3}\)

\(B=\dfrac{x^2+39+8x-24}{\left(x+3\right)\left(x-3\right)}-\dfrac{1}{x-3}\)

\(B=\dfrac{x^2+8x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}\)

\(B=\dfrac{x^2+8x+15-x-3}{\left(x+3\right)\left(x-3\right)}\)

\(B=\dfrac{x^2+7x+12}{\left(x+3\right)\left(x-3\right)}\)

\(B=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-3\right)\left(x+3\right)}\)

\(B=\dfrac{x+4}{x-3}\)

\(B=\dfrac{x+3+2\left(\sqrt{x}-3\right)+\sqrt{x}+3}{x-9}\)

\(=\dfrac{x+\sqrt{x}+6+2\sqrt{x}-6}{x-9}=\dfrac{x+3\sqrt{x}}{x-9}\)

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

\(B=\dfrac{x+3}{x-9}+\dfrac{2}{3+\sqrt{x}}-\dfrac{1}{3-\sqrt{x}}\\ B=\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\\ B=\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{x+3+2\sqrt{x}-6+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}\left(\text{đ}pcm\right)\)

ĐK: ` x\ne \pm 3`

`(x+1)/(x-3)+(x-1)/(x+3)=(x+6)/(x^2-9)`

`<=>(x+1)(x+3)+(x-1)(x-3)=x+6`

`<=>x^2+4x+3+x^2-4x+3=x+6`

`<=>2x^2+6=x+6`

`<=>2x^2-x=0`

`<=>x(2x-1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy `S={0; 1/2}`.

ĐKXĐ: x ≠ -3, x ≠ 3

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

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

\(\Rightarrow x^2+4x+3+x^2-4x+3=x+6\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

Vậy...

\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)(a)

ĐKXĐ\(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x+1\right).\left(x+3\right)+\left(x-1\right).\left(x-3\right)=x+6\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(thỏa-mãn-ĐKXĐ\right)\\x=\dfrac{1}{2}\left(thỏa-mãn-ĐKXĐ\right)\end{matrix}\right.\)

Vậy S = \(\left\{0;\dfrac{1}{2}\right\}\)

\(a,\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(dkxd:a\ne9,a\ge0\right)\)

\(=\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-3\left(\sqrt{a}-3\right)-a+2}{a-9}\)

\(=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}\)

\(=\dfrac{11}{a-9}\)

\(b,\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0,x\ne1\right)\)

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

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

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

\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)

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

\(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(\text{đ}k\text{x}\text{đ}:a\ge0;a\ne9\right)\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a-3}\right)\left(\sqrt{a+3}\right)}-\dfrac{3\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\dfrac{a-2}{\left(\sqrt{a}+3\right)\left(\sqrt{a-3}\right)}\\ =\dfrac{a+3\sqrt{a}-\left(3\sqrt{a}-9\right)-\left(a-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{11}{\left(\sqrt{a}-3\right)\left(\sqrt{a+3}\right)}\)

\(b,\dfrac{a+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\right)\\ =\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x+1}\right)}\\ =\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

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