Rút gọn biểu thức chứa căn bậc hai - Hỏi đáp

4A:

a) Ta có: \(P=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right)\cdot\frac{\sqrt{x}-2}{x}\)

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

\(=\frac{2\sqrt{x}}{\sqrt{x}+2}\cdot\frac{1}{x}\)

\(=\frac{2}{x+2\sqrt{x}}\)

4B:

a) Ta có: \(B=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

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

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

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

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

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

\(=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) Ta có: \(\frac{1}{B}=1:\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

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

Để \(\frac{1}{B}\) có giá trị nhỏ nhất thì \(\sqrt{x}+1\) nhỏ nhất

mà \(\sqrt{x}+1\ge1\forall x\ge0\)(dấu '=' xảy ra khi x=0)

nên giá trị nhỏ nhất của biểu thức \(\frac{1}{B}\) là -3 khi x=0

$ĐKXĐ : $ $ a \neq 1, $ \(a\ge0\)

Ta có :

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

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

\(=\frac{1+\sqrt{a}-a\sqrt{a}-a}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left[\left(1-\sqrt{a}\right).\left(1+\sqrt{a}\right)\right]^2}\)

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

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

\(=1\)

ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

Ta có :

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

\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2\left(1+\sqrt{a}\right)^2}\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2\left(1+\sqrt{a}\right)^2}\)

\(=\left(a+2\sqrt{a}+1\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2\left(1+\sqrt{a}\right)^2}\)

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

\(=1\)

Vậy...

Tại x=0 => B=0

Tại x khác 0 có: \(B=\frac{1}{\sqrt{x}-4+\frac{9}{\sqrt{x}}}\) (chia cả tử và mẫu cho \(\sqrt{x}\))

Áp dụng bđt cosi vs hai số dương có:

\(\sqrt{x}+\frac{9}{\sqrt{x}}\ge2\sqrt{\frac{9}{\sqrt{x}}.\sqrt{x}}\)

<=> \(\sqrt{x}-4+\frac{9}{\sqrt{x}}\ge2\sqrt{9}-4=2\)

<=> \(\frac{1}{\sqrt{x}-4+\frac{9}{\sqrt{x}}}\le\frac{1}{2}\)

<=> B\(\le\frac{1}{2}\)

Dấu "=" xảy ra <=> x=9

Lời giải:

\(\frac{\sqrt{12}-\sqrt{30}}{\sqrt{6}}.\frac{\sqrt{35}+\sqrt{42}}{\sqrt{7}}=(\frac{\sqrt{12}}{\sqrt{6}}-\frac{\sqrt{30}}{\sqrt{6}})(\frac{\sqrt{35}}{\sqrt{7}}+\frac{\sqrt{42}}{\sqrt{7}})\)

\(=(\sqrt{2}-\sqrt{5})(\sqrt{5}+\sqrt{6})=-5+2\sqrt{3}+\sqrt{10}-\sqrt{30}\)

Bài 1:

a) Ta có: \(\sqrt{\left(23-15\sqrt{3}\right)^2}\)

\(=\left|23-15\sqrt{3}\right|\)

\(=\left|\sqrt{529}-\sqrt{675}\right|\)

\(=\sqrt{675}-\sqrt{529}\)

\(=15\sqrt{3}-23\)

b) Ta có: \(\sqrt{\left(2-2\sqrt{3}\right)^2}\)

\(=\left|2-2\sqrt{3}\right|\)

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

c) Ta có: \(\sqrt{\left(15-4\sqrt{3}\right)^2}\)

\(=\left|15-4\sqrt{3}\right|\)

\(=15-4\sqrt{3}\)

d) Ta có: \(\sqrt{\left(16-6\sqrt{7}\right)^2}\)

\(=\left|16-6\sqrt{7}\right|\)

\(=\left|\sqrt{256}-\sqrt{252}\right|\)

\(=16-6\sqrt{7}\)

f) Ta có: \(\sqrt{\left(22-8\sqrt{3}\right)^2}\)

\(=\left|22-8\sqrt{3}\right|\)

\(=\left|\sqrt{484}-\sqrt{192}\right|\)

\(=22-8\sqrt{3}\)

g) Ta có: \(\sqrt{\left(9-4\sqrt{2}\right)^2}\)

\(=\left|9-4\sqrt{2}\right|\)

\(=9-4\sqrt{2}\)

h) Ta có: \(\sqrt{\left(13-4\sqrt{3}\right)^2}\)

\(=\left|13-4\sqrt{3}\right|\)

\(=13-4\sqrt{3}\)

i) Ta có: \(\sqrt{\left(7-3\sqrt{3}\right)^2}\)

\(=\left|7-3\sqrt{3}\right|\)

\(=7-3\sqrt{3}\)

a) Thay x=25 vào biểu thức \(A=\frac{7}{\sqrt{x}+8}\), ta được:

\(A=\frac{7}{\sqrt{25}+8}=\frac{7}{5+8}=\frac{7}{13}\)

Vậy: khi x=25 thì \(A=\frac{7}{13}\)

b) Ta có: \(B=\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{2\sqrt{x}-24}{x-9}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+8\sqrt{x}-3\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+8\right)-3\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)

c) Ta có: \(P=A\cdot B\)

\(=\frac{7}{\sqrt{x}+8}\cdot\frac{\sqrt{x}+8}{\sqrt{x}+3}=\frac{7}{\sqrt{x}+3}\)

ĐKXĐ: \(x\ge0\)

Để P có giá trị nguyên thì \(7⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3\inƯ\left(7\right)\)

\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-7;-1;7\right\}\)

\(\Leftrightarrow\sqrt{x}+3=7\)(vì \(\sqrt{x}+3\ge3\forall x\ge0\))

\(\Leftrightarrow\sqrt{x}=4\)

hay x=16(nhận)

Vậy: Khi x=16 thì P nguyên

d) Ta có: \(\sqrt{x}+3\ge3\forall x\ge0\)

\(\Leftrightarrow\frac{7}{\sqrt{x}+3}\le\frac{7}{3}\forall x\ge0\)

Dấu '=' xảy ra khi x=0

Vậy: Giá trị lớn nhất của biểu thức \(P=A\cdot B\) là \(\frac{7}{3}\) khi x=0

e) Để \(P=\frac{1}{2}\) thì \(\frac{7}{\sqrt{x}+3}=\frac{1}{2}\)

\(\Leftrightarrow\sqrt{x}+3=7\cdot2=14\)

\(\Leftrightarrow\sqrt{x}=14-3=11\)

hay x=121(nhận)

Vậy: để \(P=\frac{1}{2}\) thì x=121

Biểu thức không có giá trị nhỏ nhất bạn nhé.

Cứ cho $x$ từ $0$ tiến đến gần và không vượt quá $1$ thì giá trị càng âm vô cùng.

Bạn tham khảo ở câu hỏi này :

Câu hỏi của Vampire - Toán lớp 9 | Học trực tuyến

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có: \(G=\left(\frac{\sqrt{x}-2}{x-1}-\frac{2+\sqrt{x}}{1+x+2\sqrt{x}}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)

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

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

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

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

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

Để G nguyên thì \(2⋮1-x\)

\(\Leftrightarrow1-x\inƯ\left(2\right)\)

\(\Leftrightarrow1-x\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow-x\in\left\{0;-2;1;-3\right\}\)

\(\Leftrightarrow x\in\left\{0;2;-1;3\right\}\)

mà \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

nên \(x\in\left\{2;3\right\}\)

Vậy: để G nguyên thì \(x\in\left\{2;3\right\}\)

\(=\frac{\sqrt{2-\sqrt{3}}^2+\sqrt{2+\sqrt{3}}^2}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=\frac{2-\sqrt{3}+2+\sqrt{3}}{\sqrt{4-3}}=\frac{4}{1}=4\left(Do~2>\sqrt{3}\right)\)

a) Ta có: \(-B=\frac{\sqrt{x}+3}{\sqrt{x}+4}+\frac{5\sqrt{x}+12}{x-16}\)

\(\Leftrightarrow-B=\frac{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}+\frac{5\sqrt{x}+12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow-B=\frac{x-\sqrt{x}-12+5\sqrt{x}+12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow-B=\frac{x+4\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}=\frac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow-B=\frac{\sqrt{x}}{\sqrt{x}-4}\)

hay \(B=\frac{\sqrt{x}}{4-\sqrt{x}}\)

\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

\(=\dfrac{\sqrt{2}.\sqrt{4-\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{2}.\sqrt{4+\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-1}{\sqrt{2}}-\dfrac{\sqrt{7}+1}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)

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

\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

mk chép sai đề bài ở trên nha

A = \(\sqrt{7+3\sqrt{5}}+\sqrt{7-3\sqrt{5}}\)

<=> A² = ( \(\sqrt{7+3\sqrt{5}}+\sqrt{7-3\sqrt{5}}\) )²

= 7 + \(3\sqrt{5}\) + \(2\sqrt{\left(7+3\sqrt{5}\right).\left(7-3\sqrt{5}\right)}\) + 7 - \(3\sqrt{5}\)

= 14 + 2\(\sqrt{7^2-\left(3\sqrt{5}\right)^2}\)

= 14 + 2\(\sqrt{4}\)

= 18

=> A = \(\sqrt{18}\)

B = \(\sqrt{2-\sqrt{2\sqrt{5}-2}}\) - \(\sqrt{2+\sqrt{2\sqrt{5}-2}}\)

<=> B² = ( \(\sqrt{2-\sqrt{2\sqrt{5}-2}}\) - \(\sqrt{2+\sqrt{2\sqrt{5}-2}}\) )²

= 2 - \(\sqrt{2-2\sqrt{5}}\) - 2\(\sqrt{\left(2-\sqrt{2\sqrt{5}-2}\right)\left(2+\sqrt{2\sqrt{5}-2}\right)}\) + 2 + \(\sqrt{2-2\sqrt{5}}\)

= 4 - 2 \(\sqrt{2^2-\left(\sqrt{2\sqrt{5}-2}\right)^2}\)

= 4 - 2 \(\sqrt{4-\left(2\sqrt{5}-2\right)}\)

= 4 - 2 \(\sqrt{4-2\sqrt{5}+2}\)

= 4 - 2 \(\sqrt{\left(\sqrt{5}-1\right)^2}\)

= 4 - 2( \(\sqrt{5}-1\) )

= 6 - 2\(\sqrt{5}\)

=> B = \(\sqrt{6-2\sqrt{5}}\) = \(\sqrt{\left(\sqrt{5}-1\right)^2}\) = \(\sqrt{5}-1\)

Ta có: \(B=\sqrt{13+3\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+3\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}\)

\(=\sqrt{13+3\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(=\sqrt{13+3\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{13+3\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+3\cdot\left(\sqrt{2}+1\right)}\)

\(=\sqrt{13+3\sqrt{2}+3}\)

\(=\sqrt{16+3\sqrt{2}}\)

Ta có: \(C=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12}+1}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3-\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\cdot\sqrt{3-\left(\sqrt{3}+1\right)^2}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\cdot\sqrt{3-\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1}\)

\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{\sqrt{3}-1}=1\)

Ta có: \(D=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{18-\sqrt{128}}}}}\)

\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{16-2\cdot4\cdot\sqrt{2}+2}}}}\)

\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)

\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+4-\sqrt{2}}}}\)

\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4}}}\)

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

\(=\sqrt{6+2\sqrt{2}\sqrt{1}}\)

\(=\sqrt{6+2\sqrt{2}}\)

Lời giải:

Ta sẽ chứng minh PT $ax+\frac{b}{x}=c\sqrt{2}$ có nghiệm $x\neq 0$.

Với $x\neq 0$

PT $ax+\frac{b}{x}=c\sqrt{2}$

$\Leftrightarrow ax^2-c\sqrt{2}x+b=0$

$\Delta=(c\sqrt{2})^2-ab=2c^2-4ab=2[c^2-(a^2+b^2)]+2(a^2+b^2-2ab)$
$=2[c^2-(a^2+b^2)]+2(a-b)^2>0$ với mọi $c^2> a^2+b^2$
Do đó PT luôn có nghiệm.