a) \(\left(\sqrt{x}+1\right)^2\)+\(\left(\sqrt{x}+2\right)x\)+\(\left(x\sqrt{x}-\sqrt{x}+3\right)\)
b) \(\sqrt{4\left(x^2-1\right)}\)- \(2\sqrt{15}\)=0
Tìm x
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
Y
16 tháng 6 2023
\(A=3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x+6\sqrt{x}-\left(x-1\right)\)
\(=3x+6\sqrt{x}-x+1\)
\(=2x+6\sqrt{x}+1\)
\(B=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)
\(=x+3\sqrt{x}+\sqrt{x}+3-2\left(x-2\sqrt{x}+1\right)\)
\(=x+4\sqrt{x}+3-2x+4\sqrt{x}-2\)
\(=-x+8\sqrt{x}+1\)
\(C=3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x-3\sqrt{x}-2+\left(\sqrt{x^2}-1\right)\)
\(=3x-3\sqrt{x}-2+x-1\)
\(=4x-3\sqrt{x}-3\)
\(D=\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(=x-9-\left(2x-3\sqrt{x}-2\right)\)
\(=x-9-2x+3\sqrt{x}+2\)
\(=-x+3\sqrt{x}-7\)
\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)
\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)
\(=x-4-2\left(2x+3\sqrt{x}-2\right)\)
\(=x-4-4x-6\sqrt{x}+4\)
\(=-3-6\sqrt{x}\)
25 tháng 8 2021
a: Ta có: \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
\(=x-3\sqrt{x}-5\sqrt{x}-15\)
\(=x-8\sqrt{x}-15\)
b: Ta có: \(3\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
\(=3\sqrt{x}+6+2\sqrt{x}-x+6-3\sqrt{x}\)
\(=-x+2\sqrt{x}+12\)
c: Ta có: \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
\(=x-9-5\sqrt{x}+5\)
\(=x-5\sqrt{x}-4\)
d: Ta có: \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3\sqrt{x}-6-x+1\)
\(=-x+3\sqrt{x}-5\)
a) Đề có lẽ là:
đk: \(x\ge0\)
\(\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}+2\right)x=x\sqrt{x}-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}+1+x\sqrt{x}+2x-x\sqrt{x}+\sqrt{x}-3=0\)
\(\Leftrightarrow3x+3\sqrt{x}-2=0\)
\(\Leftrightarrow3\left(x+\sqrt{x}+\frac{1}{4}\right)-\frac{11}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{11}{12}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{3+\sqrt{33}}{6}\right)\left(\sqrt{x}+\frac{3-\sqrt{33}}{6}\right)=0\)
Vì \(\sqrt{x}\ge0\left(\forall x\right)\)
=> \(\sqrt{x}=\frac{3-\sqrt{33}}{6}\Rightarrow x=\frac{7-\sqrt{33}}{6}\)
b) đk: \(x\ge1\)
Ta có: \(\sqrt{4\left(x^2-1\right)}-2\sqrt{15}=0\)
\(\Leftrightarrow\sqrt{x^2-1}=\sqrt{15}\)
\(\Leftrightarrow x^2-1=15\)
\(\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)