K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

17 tháng 1 2022

a, \(P=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3\sqrt{x}}{x-9}\)

\(\Rightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow P=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}+\dfrac{3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow P=\dfrac{x-3\sqrt{x}+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow P=\dfrac{x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}\\ \Rightarrow P=\dfrac{x}{x-9}\)

b,Để P=2 \(\Leftrightarrow\dfrac{x}{x-9}=2\)

\(\Leftrightarrow x=2\left(x-9\right)\\ \Leftrightarrow x=2x-18\\ \Leftrightarrow x-18=0\\ \Leftrightarrow x=18\)

a: Khi x=16 thì \(A=\dfrac{6}{16-3\cdot4}=\dfrac{6}{4}=\dfrac{3}{2}\)

b: P=A:B

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

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

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

c: \(P-1=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}}=\dfrac{3}{\sqrt{x}}>0\)

=>P>1

 

a: Khi x=64 thì \(A=\dfrac{2}{8-2}=\dfrac{2}{6}=\dfrac{1}{3}\)

b: \(P=B:A\)

\(=\dfrac{3\sqrt{x}+\sqrt{x}-2-2\left(\sqrt{x}+2\right)}{x-4}:\dfrac{2}{\sqrt{x}-2}\)

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

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

c: P<0

=>căn x-3<0

=>0<=x<9

mà x nguyên và x<>4

nên \(x\in\left\{0;1;2;3;5;6;7;8\right\}\)

a: Khi x=121 thì \(A=\dfrac{121+3}{11+3}=\dfrac{124}{14}=\dfrac{62}{7}\)

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

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

\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

 

c: P=A:B

\(=\dfrac{x+3}{\sqrt{x}+3}:\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{x+3}{\sqrt{x}+1}\)

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

\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2>=2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}-2=2\cdot2-2=2\)

Dấu = xảy ra khi \(\left(\sqrt{x}+1\right)^2=4\)

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

=>x=1(nhận)

1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

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

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

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

2)

a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:

\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

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

=1

Thay x=1 vào P, ta được:

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

17 tháng 5 2021

1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)

Thay \(x=\frac{1}{9}\) vào A ta có:

\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)

2. \(B=...\)

    \(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

    \(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{\sqrt{x}+3}{-6}\)

Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)

hay \(P\le-\frac{1}{2}\)

Dấu "=" xảy ra <=> x=0

17 tháng 5 2021

toán lớp 9 khó zậy em đọc k hỉu 1 phân số

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

\(M=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b: \(A=\dfrac{-3x+4x+7}{\sqrt{x}+3}=\dfrac{x+7}{\sqrt{x}+3}=\dfrac{x-9+16}{\sqrt{x}+3}\)

=>\(A=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)

 

Dấu = xảy ra khi x=1

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

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

b: A=1/3

=>\(\dfrac{-3}{\sqrt{x}-3}=\dfrac{1}{3}\)

=>căn x-3=-9

=>căn x=-6(loại)

c: căn x-3>=-3

=>3/căn x-3<=-1

=>-3/căn x-3>=1

Dấu = xảy ra khi x=0

11 tháng 8 2023

\(-3+6=-3\) =))

30 tháng 9 2018

Giúp tớ nhanh nhanh nha!Cảm ơn rất rất nhiều.

AH
Akai Haruma
Giáo viên
28 tháng 12 2023

Lời giải:

a.

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

b.

\(P=AB=\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}+3}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

Để $P<0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+3}<0$

Mà $\sqrt{x}+3>0$ nên $\sqrt{x}-2<0$

$\Leftrightarrow 0< x< 4$

Kết hợp với ĐKXĐ suy ra $0< x< 4$

Mà $x$ nguyên nên $x\in left\{1; 2; 3\right\}$