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

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

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

\(=\dfrac{2x+6}{2x\left(x+3\right)}\)

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

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

Vậy \(A=x\)

b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)

Vậy...

2/a,

\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)

\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)

\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)

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

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

Vậy....

b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)

Vậy..

câu 2 làm tương tự câu 1 nha

Kiếm việc làm nào :)

1) ĐK \(x\ne\pm9\)

\(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x+9}{x-9}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\frac{3x+9}{x-9}\)

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

2) ?

3) Ta có

\(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\)

\(\Rightarrow A=\frac{3}{\sqrt{x}+3}\le1\)

Dấu "=" xảy ra khi x = 0

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

\(=\sqrt{x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2}\)

\(=\sqrt{\left(x-\frac{1}{2}\right)^2}\)

\(=\left|x-\frac{1}{2}\right|\)

2: Ta có: \(\sqrt{x^2}+\sqrt{x^6}\)

\(=\sqrt{x^2}\cdot1+\sqrt{x^2}\cdot\sqrt{x^4}\)

\(=\sqrt{x^2}\cdot\left(1+\sqrt{x^4}\right)\)

\(=\left|x\right|\cdot\left(1+x^2\right)\)

3: Ta có: \(C=\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)

\(=\sqrt{2-2\cdot\sqrt{2}\cdot1+1}-\sqrt{4-2\cdot2\cdot\sqrt{2}+2}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\left|\sqrt{2}-1\right|-\left|2-\sqrt{2}\right|\)

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

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

:v Thay cái câu đó = mấy cái dấu roài giải BPT thôi mà

Nếu dễ vậy sao bạn ko làm cho rõ đi định quang

mk làm xong hết rối k cần nữa đâu

a)P\(=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\)

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

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

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

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

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

bài 1: a) \(A=\frac{\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right)}{\frac{a+2}{a-2}}\)

\(A=\left(\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{a+2}{a-2}\)

\(A=\left(\frac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\right)\cdot\frac{a-2}{a+2}\)

\(A=2\cdot\frac{a-2}{a+2}\left(a\ne0;a\ne\pm2\right)\)

b) để A = 1 => \(2\cdot\frac{a-2}{a+2}=1\)

=> 2a - 4 = a + 2

=> a = 6 (thỏa mãn)

bài 2) a) ĐKXĐ: \(x\ne4\)

b) \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(\Leftrightarrow B=\frac{2\sqrt{x}+\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\frac{2\sqrt{x}+4}{x-4}=\frac{2}{\sqrt{x}-2}\)

c) \(B=\frac{2}{\sqrt{3+2\sqrt{3}}-2}\) \(\approx3,69\)

(bạn tự bấm máy tính nhé nhưng theo mình thấy nếu x = 4 + 2\(\sqrt{3}\) hay \(3+2\sqrt{2}\) thì sẽ cho kết quả đẹp hơn, k biết bạn có nhầm đề k nữa!)

bài 3)

a, \(B=\left(\frac{1}{\sqrt{y}+1}-\frac{3\sqrt{y}}{\sqrt{y}-1}+3\right)\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\left(y\ne1;y\ne4\right)\)

\(\Leftrightarrow B=\frac{\sqrt{y}-1-3y-3\sqrt{y}+3y-3}{y-1}\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\)

\(\Rightarrow B=\frac{-2\sqrt{y}-4}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\Rightarrow B=\frac{-2}{\sqrt{y}-1}\)

b) y = \(3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)

=> B = \(\frac{-2}{\sqrt{\left(1+\sqrt{2}\right)^2}-1}\)

\(\Rightarrow B=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

a/ Chia đa thức một biến bình thường. Ta sẽ có thương là n² - 1, số dư là 7

Để n³ +n²-n+5 chia hết cho n+2

thì 7 chia hết cho n+2

\(\Rightarrow\)n+2\(_{ }\in\)Ư(7)

\(\Rightarrow\)n+2\(\in\)\(\left\{1,-1,7,-7\right\}\)

\(\Rightarrow n\in\left\{-1,-3,5,-9\right\}\)

Câu b tương tự

Các bn giúp mk vs nha ! Mk cảm ơn trước.