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Lời giải:
Đặt \(\sqrt{x}=a(a\geq 0)\)
Khi đó:\(M=\frac{a^3-4a^2-a+4}{2a^3-14a^2+28a-16}\)
a) Điều kiện để M có nghĩa:
\(2a^3-14a^2+28a-16\neq 0\Leftrightarrow 2a^2(a-1)-12a(a-1)+16(a-1)\neq 0\)
\(\Leftrightarrow (a-1)(2a^2-12a+16)\neq 0\)
\(\Leftrightarrow (a-1)[2a(a-4)-4(a-4)]\neq 0\)
\(\Leftrightarrow 2(a-1)(a-2)(a-4)\neq 0\Leftrightarrow a\neq 1; a\neq 2; a\neq 4\)
Suy ra điều kiện để M có nghĩa là $x\geq 0; x\neq 1; x\neq 4; x\neq 16$
b)
\(M=\frac{a^3-4a^2-a+4}{2a^3-14a^2+28a-16}=\frac{a^2(a-4)-(a-4)}{2(a-1)(a-2)(a-4)}\)
\(=\frac{(a^2-1)(a-4)}{2(a-1)(a-2)(a-4)}=\frac{(a-1)(a+1)(a-4)}{2(a-1)(a-2)(a-4)}=\frac{a+1}{2(a-2)}=\frac{\sqrt{x}+1}{2(\sqrt{x}-2)}\)
c)
Để M nhận giá trị nguyên thì \(\sqrt{x}+1\vdots 2(\sqrt{x}-2)\)
\(\Rightarrow \sqrt{x}+1\vdots \sqrt{x}-2\)
\(\Rightarrow \sqrt{x}-2+3\vdots \sqrt{x}-2\)
\(\Rightarrow 3\vdots \sqrt{x}-2\Rightarrow \sqrt{x}-2\in\left\{\pm 1;\pm 3\right\}\)
\(\Rightarrow \sqrt{x}\in\left\{1; 3; 5\right\}\Rightarrow x\in\left\{1;9;25\right\}\)
Thử lại thấy đều thỏa mãn.
a/ ĐKXĐ: \(x\ge0;x\ne1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]-1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\dfrac{\left(x+1+\sqrt{x}\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+1+\sqrt{x}}{\sqrt{x}-1}-1=\dfrac{x+2}{\sqrt{x}-1}\)
b/ Ta có:
\(Q=P-\sqrt{x}\)
= \(\dfrac{x+2}{\sqrt{x}-1}-\sqrt{x}\)
= \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\)
Để Q nhận giá trị nguyên thì \(1+\dfrac{3}{\sqrt{x}-1}\in Z\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}-1}\in Z\) ( vì 1\(\in Z\) )
\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(3\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=3\\\sqrt{x}-1=-3\\\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=-2\\\sqrt{x}=2\\\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\left(tm\right)\\\\x=4\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì x=\(\left\{16;4;0\right\}\)
a) ĐKXĐ: x ≥ 0, x # 9
(quá trình thì b tự làm nha)
b) Ta có:
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{7\sqrt{x}+4}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
= \(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+7\sqrt{x}+4-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x-3}\right)}\)
= \(\dfrac{2x-3\sqrt{x}+7\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
= \(\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
Vậy A = \(\dfrac{\sqrt{x}}{\sqrt{x}+2}\) với x ≥ 0, x # 9
c) ĐKXĐ: x ≥ 0, x # 9, x ∈ Z
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-2}{\sqrt{x}+2}=1-\dfrac{2}{\sqrt{x}+2}\)
Để A có gtn ⇔ \(1-\dfrac{2}{\sqrt{x}+2}\) nguyên ⇔ \(\dfrac{2}{\sqrt{x}+2}\) nguyên
⇔ \(\sqrt{x}+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
⇒ \(\sqrt{x}+2\in\left\{1;2\right\}\) (Vì \(\sqrt{x}+2>0\))
Nếu \(\sqrt{x}+2=1\)thì \(\sqrt{x}=-1\) (Vô nghiệm)
Nếu \(\sqrt{x}+2=2\) thì \(\sqrt{x}=0\Leftrightarrow x=0\)(TMĐK)
Vậy để A có gtn thì x = 0
a:ĐKXĐ: x>=0; \(x\notin\left\{4;9\right\}\)
\(A=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{2x-4\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-x-2x+3\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3x+5\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3x+6\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3\sqrt{x}-1}{\sqrt{x}-3}\)
b: Để A là số nguyên thì \(-3\sqrt{x}+9-10⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(x\in\left\{16;25;1;64;169\right\}\)
a) \(A=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\dfrac{2}{\sqrt{x}+2}=1+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=1+\dfrac{2\left(\sqrt{x}-2\right)}{x-4}\)Thay x = 6 vào A ta có :
\(A=1+\dfrac{2\left(\sqrt{6}-2\right)}{6-4}=1+\sqrt{6}-2=\sqrt{6}-1\)
b)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{x+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{\sqrt{x}+2}{x-16}\)
a)
\(P=\dfrac{\left|\sqrt{x+4\sqrt{x-4}+\sqrt{x-4\sqrt{x-4}}}\right|}{\sqrt{1-\dfrac{8}{x}+\dfrac{16}{x^2}}}\)
\(=\dfrac{\sqrt{x-4+2.2\sqrt{x-4}+2^2}+\sqrt{x-4-2.2\sqrt{x-4}+2^2}}{\sqrt{\left(\dfrac{4}{x}\right)^2-\dfrac{2.4}{x}+1^2}}\)
\(=\dfrac{\left(\sqrt{x-4}+2\right)^2+\left(\sqrt{x-4}-2\right)^2}{\sqrt{\left(\dfrac{4}{x}-1\right)^2}}\)
Tự giải tiếp nha :)
a: ĐKXĐ: x>1
b: \(B=\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}+x=x+2\sqrt{x-1}\)
c: Khi B=4 thì \(\left(\sqrt{x-1}+1\right)^2=4\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
=>x-1=1
=>x=2
a: \(A=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b: \(A=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu '=' xảy ra khi x=1/4
1) Khi x = 36 thì A = \(\frac{\sqrt{36}+4}{\sqrt{36}+2}\Leftrightarrow\frac{5}{4}\)
Vậy khi x = 36 thì A = \(\frac{5}{4}\)
2) B = \((\frac{\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\frac{4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}):\frac{x+16}{\sqrt{x}+2}\)
= \(\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\frac{\sqrt{x}+2}{x+16}=\frac{x+16}{x-16}.\frac{\sqrt{x}+2}{x+16}\)
= \(\frac{\sqrt{x}+2}{x-16}\)
Vậy B = \(\frac{\sqrt{x}+2}{x-16}\)
ĐK: x>0,\(x\ne1\)
a) \(Q=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2x-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
Ta có Q=\(x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow Q\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy GTNN của Q là \(\dfrac{3}{4}\) và xảy ra khi \(x=\dfrac{1}{4}\)
b)
Ta có \(\dfrac{3Q}{\sqrt{x}}=\dfrac{3\left(x-\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{3x-3\sqrt{x}+3}{\sqrt{x}}=3\sqrt{x}-3+\dfrac{3}{\sqrt{x}}\)Vậy để \(\dfrac{3Q}{\sqrt{x}}\) nguyên thì \(\left\{{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\inƯ\left(3\right)\in\left(1;3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=9\left(tm\right)\end{matrix}\right.\)
Vậy x=9 thì \(\dfrac{3Q}{\sqrt{x}}\) nhận giá trị nguyên
Lời giải:
a.
\(A=\frac{(x\sqrt{x}-4x)-(\sqrt{x}-4)}{2(\sqrt{x}-4)(\sqrt{x}-2)(\sqrt{x}-1)}\)
ĐKXĐ: \(\left\{\begin{matrix} x\geq 0\\ \sqrt{x}-4\neq 0\\ \sqrt{x}-2\neq 0\\ \sqrt{x}-1\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\neq 16\\ x\neq 4\\ x\neq 1\end{matrix}\right.\)
\(A=\frac{x(\sqrt{x}-4)-(\sqrt{x}-4)}{2(\sqrt{x}-4)(\sqrt{2}-2)(\sqrt{x}-1)}=\frac{(x-1)(\sqrt{x}-4)}{2(\sqrt{x}-4)(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(=\frac{(\sqrt{x}-1)(\sqrt{x}+1)(\sqrt{x}-4)}{2(\sqrt{x}-4)(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{\sqrt{x}+1}{2(\sqrt{x}-2)}\)
b.
Với $x$ nguyên, để $A\in\mathbb{Z}$ thì $\sqrt{x}+1\vdots 2(\sqrt{x}-2)}$
$\Rightarrow \sqrt{x}+1\vdots \sqrt{x}-2$
$\Leftrightarrow \sqrt{x}-2+3\vdots \sqrt{x}-2$
$\Leftrightarrow 3\vdots \sqrt{x}-2$
$\Rightarrow \sqrt{x}-2\in\left\{\pm 1;\pm 3\right\}$
$\Rightarrow x\in\left\{1;9;25\right\}$
Thử lại thấy đều thỏa mãn.