Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
a) ta thấy x-4=(canx-2)(cãnx+2)
2-canx=-(cãnx - 2)
tự học mới giỏi
b)rut gọn roi giai cho
Giải
Đặt A = √x2+11x−6−3√x+6
B = √x2+3x−2−3√x+2
Theo bài ra ta có A + B = 4 (1)
Mặt khác ta có A2 - B2 = 8x + 32 - 24√2x−1(2)
Từ (1) ta có A = 4 - B thế vào (2) ta có 16 - 8B + B2 - B2 = 8x + 32 - 24√2x−1
Hay B + x + 2 - 3√2x−1= 0⇒√x2+3x−2−3√x+2+x+2 - 3√2x−1⇒√(x+2)(2x−1) - 3√2x−1+√x+2(√x+2−3)= 0
Hay √2x−1(√x+2−3)+√x+2(√x+2−3)=0
⇒(√x+2−3)(√2x−1+√x+2)=0
⇔√x+2−3=0⇔x=7
Thử lại x = 7 thỏa mã bài ra. Vậy nghiệm của phương trình la x = 7
Đúng 7 Hưng đã chọn câu trả lời này.
Câu a bạn tự làm nha!. Câu b : A=\(\frac{2x}{x-1}\)=\(\frac{2x-2}{x-1}\)-\(\frac{2}{x-1}\)=\(\frac{2.\left(x-1\right)}{x-1}\)-\(\frac{2}{x-1}\)=2-\(\frac{2}{x-1}\). Để A nguyên thì x-1 là ước của 2. Đến đó dễ rồi bạn tự làm nha. Học tốt!
a) ĐKXĐ: \(x\ne4\)và \(x>0\)
............................
\(\Leftrightarrow A=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}+2}\right)\)\(:\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{3x-6\sqrt{x}\left(\sqrt{x}+2\right)+3\sqrt{x}\left(\sqrt{x}-2\right)}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}:\left(\frac{x-2+10-x}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{3x-6x-12\sqrt{x}+3x-6\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\left(\frac{8}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{-18\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}-2}{8}\)
\(\Leftrightarrow A=\frac{-3}{4\left(\sqrt{x}+2\right)}\)
Vậy \(A=\frac{-3}{4\left(\sqrt{x}-2\right)}\)với \(x>0\)và \(x\ne4\)
b)Ta có \(A< 2\Leftrightarrow\frac{-3}{4\left(\sqrt{x}-2\right)}< 2\)
\(\Leftrightarrow\frac{-3}{4\left(\sqrt{x}-2\right)}-2< 0\)
\(\Leftrightarrow\frac{-3-8\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{-3-8\sqrt{x}-16}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{-18-8\sqrt{x}}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow-18-8\sqrt{x}< 0\)( Vì \(4\left(\sqrt{x}-2\right)>0\)với \(\forall x\))
\(\Leftrightarrow\sqrt{x}< \frac{-9}{4}\)(Vô Nghiệm)
Vậy không có gtr nào của x thỏa mãn A<2
a) \(A=\left(\sqrt{6}+\sqrt{10}\right).\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{2}\)
\(B=\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}+1\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+1\)
\(=\frac{4}{x-4}+1\)
\(=\frac{4}{x-4}+\frac{x-4}{x-4}=\frac{x}{x-4}\)
lam gjup vs mn oi
1/ ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
\(A=\left[\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(=\left[\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{6}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)}{6}\)
\(=\frac{-2}{\sqrt{x}-2}.\frac{1}{6}=-\frac{1}{3\left(\sqrt{x}-2\right)}\)
2/ Để \(A>2\Rightarrow\frac{-1}{3\left(\sqrt{x}-2\right)}>2\)\(\Rightarrow6\sqrt{x}-12+1>0\Rightarrow6\sqrt{x}-11>0\Rightarrow\sqrt{x}>\frac{11}{6}\)
\(\Rightarrow x>\frac{121}{36}\)