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a) Đặt \(\sqrt{x}=a\Rightarrow B=\left(\dfrac{a}{a+4}+\dfrac{4}{a-4}\right):\dfrac{a^2+16}{a+2}\)
Quy đồng,rút gọn : \(B=\dfrac{a+2}{a^2-16}\Rightarrow B=\dfrac{\sqrt{x}+2}{x-16}\)
b) \(B\left(A-1\right)=\dfrac{\sqrt{x}+2}{x-16}\left(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)=\dfrac{2}{x-16}\)
x - 16 là ước của 2 => \(x\in\left\{14;15;17;18\right\}\)
mới làm quen toán 9 ;v có gì k rõ ae chỉ bảo nhé :))
\(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\left(ĐKXĐ:x\ne1;x\ge0\right)\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x+3}}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-8+5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-3\sqrt{x}+8\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)
b)Để \(P< \frac{15}{4}\)thì \(\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}< \frac{15}{4}\)
Ta có:\(\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}< \frac{15}{4}\)
\(\Leftrightarrow\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}-\frac{15}{4}< 0\)
\(\Leftrightarrow\frac{12\sqrt{x}+32-15\sqrt{x}-30}{4\left(\sqrt{x}+2\right)}< 0\)
\(\Leftrightarrow\frac{-\left(3\sqrt{x}+2\right)}{4\sqrt{x}+8}< 0\)
Vì \(x\ge0;x\ne1\)
Do đó \(0< 4\sqrt{x}+8\)
Mà \(-\left(3\sqrt{x}+2\right)< 0\)
Vậy \(P< \frac{15}{4}\left(đpcm\right)\)
c)Ta có:\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow P=\frac{3\sqrt{x}+6+2}{\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow P=\frac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}+\frac{2}{2\sqrt{x}+2}\)
\(\Leftrightarrow P=3+\frac{2}{\sqrt{x}+2}\)
Vì \(x\ge0;x\ne1\Rightarrow\frac{2}{\sqrt{x}+2}\le1\)
Do đó \(P\le4\Leftrightarrow x=1\)
Vậy Max P=4 khi x=1
P=3x+3√x−9(√x−1)(√x+2) +√x+3√x+2 −√x−2√x−1
P=3x+3√x−9(√x−1)(√x+2) +(√x+3)(√x−1)(√x+2)(√x−1) −x−4(√x−1)(√x+2)
P=3x+3√x−9+x+2√x−3−x+4(√x−1)(√x+2)
P=3x−8+5√x(√x−1)(√x+2)
P=3x−3√x+8√x−8(√x−1)(√x+2)
P=(3√x+8)(√x−1)(√x−1)(√x+2)
P=(3√x+8)(√x+2)
b)Để P<154 thì (3√x+8)(√x+2) <154
Ta có:(3√x+8)(√x+2) <154
⇔(3√x+8)(√x+2) −154 <0
⇔12√x+32−15√x−304(√x+2) <0
⇔−(3√x+2)4√x+8 <0
Vì x≥0;x≠1
Do đó 0<4√x+8
Mà −(3√x+2)<0
Vậy P<154 (đpcm)
c)Ta có:P=(3√x+8)(√x+2)
⇔P=3√x+6+2(√x+2)
⇔P=3(√x+2)(√x+2) +22√x+2
⇔P=3+2√x+2
Vì x≥0;x≠1⇒2√x+2 ≤1
Do đó
ĐKXĐ: \(x\ge0\)
a/ Đề \(=\left(\frac{1-\sqrt{x}^3}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left[\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right]\left[\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right]\)
\(=\left(1+2\sqrt{x}+x\right)\left(1-2\sqrt{x}+x\right)\)
\(=\left(1+\sqrt{x}\right)^2\left(1-\sqrt{x}\right)^2\)
\(=\left[\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\right]^2=\left(1-x\right)^2\)
b/ \(P< 7-4\sqrt{3}\Leftrightarrow\left(1-x\right)^2< 7-4\sqrt{3}\)
\(\Rightarrow\left(1-x\right)^2< \left(2-\sqrt{3}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-x< 2-\sqrt{3}\Rightarrow x>-1+\sqrt{3}\\1-x< \sqrt{3}-2\Rightarrow x>3-\sqrt{3}\end{cases}}\)
Vậy \(x>3-\sqrt{3}\)