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a. ĐKXĐ : x>1.
b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)
c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:
\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)
Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).
a ) ĐK : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)\(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^{^2}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{x+4\sqrt{x}+3}\)
A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
Ta có:
A = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
= \(\dfrac{-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-1\)
Kết luận: ...
ĐK của nó còn là: x ≥ 0 nữa dung doan nhé, mình viết thiếu...
Lần sau ghi dấu ra xíu nhé :v
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 :))
Bài 2:
a) Thay m=3 vào hệ pt, ta được:
\(\left\{{}\begin{matrix}x-2y=7\\2x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=14\\2x+y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5y=5\\x-2y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=7+2y=5\end{matrix}\right.\)
Vậy: Khi m=3 thì hệ phương trình có nghiệm duy nhất là (x,y)=(5;-1)
a) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{4}{\sqrt{x}+1}\)
b. Để A và B trái dấu \(\Leftrightarrow AB< 0\)
\(\Leftrightarrow\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{4}{\sqrt{x}+1}\right)< 0\)
\(\Leftrightarrow\dfrac{4}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\)
\(\Rightarrow0< x< 1\)
ĐKXĐ: \(x\ge0;x\ne1\)
Sửa lại đề chỗ \(\dfrac{\sqrt{x-1}}{\sqrt{x}+2}\) thành \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(P=\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=2-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{2\sqrt{x}+4-\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}=1+\dfrac{3}{\sqrt{x}+2}\)
Để P lớn nhất \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\) lớn nhất
Mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)
\(\Rightarrow P_{max}=1+\dfrac{3}{2}=\dfrac{5}{2}\) khi \(\sqrt{x}+2=2\Leftrightarrow x=0\)
Khôi Bùi , DƯƠNG PHAN KHÁNH DƯƠNG, Mysterious Person, Phạm Hoàng Giang, Phùng Khánh Linh, TRẦN MINH HOÀNG, Dũng Nguyễn, Nhã Doanh, hattori heiji, ...
a: \(A=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(x+1\right)\left(y+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
b: \(x=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
\(A=\dfrac{2\sqrt{\sqrt{2}-1}}{\sqrt{2}-1+1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
bn ơi! sai đề thì pải
a: \(C=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{1}{x-\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}+1-3+\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\dfrac{x-1}{x\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
b: Để C<1 thì C-1<0
\(\Leftrightarrow\dfrac{\sqrt{x}-1-x+\sqrt{x}-1}{x-\sqrt{x}+1}< 0\)
=>\(-x+2\sqrt{x}-2< 0\)(luôn đúng)