Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
cho S=1-3+32-33+...+398-399
a. Chứng minh: S chia hêt cho 20
b. Rút gọn S, từ đó suy ra 3100 chia 4 dư 1
chịu
\(A=\frac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4}{x-1}\)
b) \(\frac{4}{x-1}=7\)
\(\Leftrightarrow4=7.\left(x-1\right)\)
\(\Leftrightarrow\frac{4}{7}=x-1\)
\(\Leftrightarrow\frac{4}{7}+1=x\)
\(\Leftrightarrow\frac{11}{7}=x\)
\(\Rightarrow x=\frac{11}{7}\)
ĐK :\(\hept{\begin{cases}x>=0\\x\ne1\end{cases}}\)
Ta có: \(A=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)+x-1}\right]:\left[\frac{\sqrt{x}+1}{x-1}-\frac{2}{x-1}\right]\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\frac{1}{\sqrt{x}-1}\right).\left(\frac{x+1}{x+1+\sqrt{x}}\right)\)
\(=\frac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\frac{x+1}{x+\sqrt{x}+1}=\frac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}.\frac{1}{x+\sqrt{x}+1}=\frac{-\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}\)
\(\text{ĐKXĐ: }x>0\)
\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+1=\frac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{x-\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\)
\(=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)
\(\text{Vậy }P=x-\sqrt{x}=x-2\sqrt{x}\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\left(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\text{Với mọi x}\right)\)
\(\text{Dấu "=" xảy ra khi: }\sqrt{x}-\frac{1}{2}=0\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{4}\)
\(\text{Vậy GTNN của P là : }-\frac{1}{4}\text{ tại : }x=\frac{1}{4}\)
sao biểu thức khi rút gọn xấu vậy bạn ? đề có sai khum :vv, thế tìm x dài lắm bạn ạ
a, Với x > 0 ; \(x\ne1\)
\(M=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}-x}\right)\)
\(=\left(\frac{x+\sqrt{x}+x-\sqrt{x}}{x-1}\right):\left(\frac{2\sqrt{x}-2-2+x}{x\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{2x}{x-1}\right):\left(\frac{x+2\sqrt{x}-4}{x\left(\sqrt{x}-1\right)}\right)=\frac{2x^2}{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}-4\right)}\)
a, \(P=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+1\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}-1+1=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)
b, \(P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\)
Vậy Min P =-1/4
c, Chắc bằng nhau vì cùng dương mà
Phần a như bạn Đỗ Ngọc Hải chỉ thêm ĐKXĐ : x >= 0
b) Đkxd X >=0
Ta Có P = x-\(\sqrt{x}\) -2√x.½+1/4 -1/4=\(\left(\sqrt{x}-\frac{1}{2}\right)^2\)\(-\frac{1}{4}\)
Có √x>=0<=> (√x-½)2>=1/4<=>(√x-½)2-1/4>=0=>P>=0
Hay min p =0
Dấu = xảy ra <=> x=0
Vậy để minP=0<=>x=0
C)Dkxd x>1
CóP>=0(chứng minh trên )
=>|P|=P