Tìm đk và rút gọn , tìm x để A < 0 , tính A biết x =...
Đọc tiếp
Tìm đk và rút gọn , tìm x để A < 0 , tính A biết x = 3-2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
VT
1
H9
HT.Phong (9A5)
CTVHS
15 tháng 9 2023
\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\) (ĐK: \(x>0;x\ne1;x\ne9\))
\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1+x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2x-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{2\left(x-5\right)}\)
\(=\dfrac{\sqrt{x}-3}{2\sqrt{x}\left(x-5\right)}\)
\(=\dfrac{\sqrt{x}-3}{2x\sqrt{x}-10\sqrt{x}}\)
\(A>0\) khi
\(\dfrac{\sqrt{x}-3}{2x\sqrt{x}-10\sqrt{x}}>0\)
TH1:
\(\sqrt{x}-3>0\) và \(2x\sqrt{x}-10\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}>3\) và \(2\sqrt{x}\left(x-5\right)>0\)
\(\Leftrightarrow x>9\) và \(x>5\)
\(\Leftrightarrow x>9\)
TH2:
\(\sqrt{x}-3< 0\) và \(2x\sqrt{x}-10\sqrt{x}< 0\)
\(\Leftrightarrow\sqrt{x}< 3\) và \(2\sqrt{x}\left(x-5\right)< 0\)
\(\Leftrightarrow x< 9\) và \(x< 5\)
\(\Leftrightarrow x< 5\)
Vậy A > 0 khi \(\left[{}\begin{matrix}x>9\\x< 5\end{matrix}\right.\)
Ta có:
\(x=3-2\sqrt{2}=\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2=\left(\sqrt{2}-1\right)^2\)
\(A=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-3}{2\cdot\left(\sqrt{2}-1\right)^2\cdot\sqrt{\left(\sqrt{2}-1\right)^2}-10\cdot\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(A=\dfrac{\left|\sqrt{2}-1\right|-3}{2\cdot\left(3-2\sqrt{2}\right)\cdot\left|\sqrt{2}-1\right|-10\cdot\left|\sqrt{2}-1\right|}\)
\(A=\dfrac{\sqrt{2}-1-3}{\left(6-4\sqrt{2}\right)\left(\sqrt{2}-1\right)-10\left(\sqrt{2}-1\right)}\)
\(A=\dfrac{\sqrt{2}-4}{6\sqrt{2}-6-8+4\sqrt{2}-10\sqrt{2}+10}\)
\(A=\dfrac{\sqrt{2}-4}{-4}\)
\(A=\dfrac{4-\sqrt{2}}{4}\)
23 tháng 12 2020
a) Điều kiện: \(x\ne\pm1\)
\(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)
\(B=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{-4}{\left(x-1\right).\left(x+1\right)}\)
\(B=\frac{x^2-x-x+1-x^2-x-x-1+4}{\left(x-1\right).\left(x+1\right)}\)
\(B=\frac{-4x+4}{\left(x-1\right).\left(x+1\right)}=\frac{-4.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}=\frac{-4}{x+1}\)
b) \(x^2-x=0\Leftrightarrow x.\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Khi \(x=0\Leftrightarrow\frac{-4}{0-1}=\frac{-4}{-1}=4\)
Khi \(x=1\Leftrightarrow\frac{-4}{1-1}=0\)
c) \(\frac{-4}{x+1}=-3\Leftrightarrow-3.\left(x+1\right)=-4\Leftrightarrow x+1=\frac{4}{3}\Leftrightarrow x=\frac{1}{3}\)
30 tháng 12 2021
1: \(B=\dfrac{6x+x^2-3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x}{x-3}\)
18 tháng 12 2023
Bài 1:
a: Sửa đề \(x^3y-2x^2y+xy\)
\(=y\left(x^3-2x^2+x\right)\)
\(=x\cdot y\cdot\left(x^2-2x+1\right)\)
\(=xy\left(x-1\right)^2\)
b: Sửa đề: \(x^2-9-2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)-9\)
\(=\left(x-y\right)^2-9\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
Bài 2:
a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)
b: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)
\(=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{2x+6-x-5}{x+3}\)
\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)
\(=\dfrac{x^2-3x-2x-6-x^2+1}{x-3}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{-5x-5}{\left(x-3\right)\left(x+1\right)}=-\dfrac{5\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}=-\dfrac{5}{x-3}\)
c: \(x^2-x-2=0\)
=>\(\left(x-2\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Thay x=2 vào A, ta được:
\(A=\dfrac{-5}{2-3}=\dfrac{-5}{-1}=5\)
19 tháng 2 2023
a: \(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-2x}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
b: \(P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
Để |P|>P thì P<0
=>căn x-2<0
=>0<x<4
=>x=1
16 tháng 11 2021
\(a,ĐK:x\ne3;x\ge1\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\\ b,A=4\left(2-\sqrt{3}\right)\\ \Leftrightarrow\sqrt{x-1}+\sqrt{2}=8-4\sqrt{3}\\ \Leftrightarrow\sqrt{x-1}=8-4\sqrt{3}-\sqrt{2}\\ \Leftrightarrow x-1=\left(8-4\sqrt{3}-\sqrt{2}\right)^2\\ \Leftrightarrow x=\left(8-4\sqrt{3}-\sqrt{2}\right)^2+1=...\\ d,A=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Lời giải:
ĐKXĐ: $x>0; x\neq 1; x\neq 9$
\(A=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)+(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}-1)}\)
\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{x-1-(x-9)}{(\sqrt{x}-3)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{8}{(\sqrt{x}-1)(\sqrt{x}-3)}\)
\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)(\sqrt{x}-3)}{8}=\frac{\sqrt{x}-3}{8\sqrt{x}}\)
Để $A<0\Leftrightarrow \frac{\sqrt{x}-3}{8\sqrt{x}}<0$
$\Leftrightarrow \sqrt{x}-3<0$ (do $8\sqrt{x}>0$)
$\Leftrightarrow \sqrt{x}<3$
$\Leftrightarrow 0\leq x< 9$
Kết hợp với đkxđ suy ra $0< x< 9; x\neq 1$
Khi $x=3-2\sqrt{2}=(\sqrt{2}-1)^2$
$\Rightarrow \sqrt{x}=\sqrt{2}-1$
Khi đó: $A=\frac{\sqrt{x}-3}{8\sqrt{x}}=\frac{\sqrt{2}-4}{8(\sqrt{2}-1)}=\frac{-2-3\sqrt{2}}{8}$
sao chỗ x−1−(x−9) lại là trừ ạ đáng lẽ nó phải là (x−1)+(x−9) chứ ạ