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1) Liên hợp hay bình phương gì gì cx được nếu bạn rảnh =))
2)Giải PT : $5^{x}= 3^{x}+ 4^{x}$ - Các bài toán và vấn đề về PT - HPT - BPT - Diễn đàn Toán học
4) Câu hỏi của VanCan - Toán lớp 8 - Học toán với OnlineMath
Mấy bài này dài vật vã ghê =)))))))))))))
1, a, \(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-5}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{8+4\sqrt{3}-5}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}\)
=\(\sqrt{6}+\sqrt{2}+\sqrt{5}\)
b, M = \(\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x^2}-x+1}\)(ĐKXĐ: \(x\ge0\))
= \(\frac{\sqrt{3}\left(x-1\right)}{x-x+1}\)
= \(\sqrt{3}\left(x-1\right)\)
Thay x = \(2+\sqrt{3}\)(TMĐK) vào M ta có:
M = \(\sqrt{3}\left(2+\sqrt{3}-1\right)=\sqrt{3}\left(1+\sqrt{3}\right)=3+\sqrt{3}\)
Vậy với x = \(2+\sqrt{3}\)thì M = \(3+\sqrt{3}\)
2, Mình chỉ giải câu a thôi nhé:
\(\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt{1+a}\)
\(\Leftrightarrow\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge\left(2\sqrt{1+a}\right)^2\)
\(\Leftrightarrow1+b+2\sqrt{\left(1+b\right)\left(1+c\right)}+1+c\ge4\left(1+a\right)\)
\(\Leftrightarrow2+b+c+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)\left(1\right)\)
Vì \(\left(\sqrt{1+b}-\sqrt{1+c}\right)^2\ge0\)
\(\Rightarrow2+b+c\ge2\sqrt{\left(1+b\right)\left(1+c\right)}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4+2\left(b+c\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4\left(1+a\right)\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4+4a\)
\(\Leftrightarrow2\left(b+c\right)\ge4a\)
\(\Leftrightarrow b+c\ge2a\)
4*. Thật ra cái này mình xài làm trội, làm giảm là được mà
Đặt A = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+....+\frac{1}{2\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+....+\frac{1}{\sqrt{n}+\sqrt{n}}\)
Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2}}>\frac{1}{\sqrt{3}+\sqrt{2}}\)
\(\frac{1}{\sqrt{3}+\sqrt{3}}>\frac{1}{\sqrt{4}+\sqrt{3}}\)
+ .........................................................
\(\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
Cộng tất cả vào
\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{n+1}+\sqrt{n}}\)\(\frac{1}{2}A>\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
\(\frac{1}{2}A>\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\)
\(\frac{1}{2}A>\sqrt{n+1}-\sqrt{2}\)
\(A>2\sqrt{n+1}-2\sqrt{2}>2\sqrt{n+1}-3\)
\(A+1>2\sqrt{n+1}-3+1\)
\(A+1>2\sqrt{n+1}-2\)
\(A+1>2\left(\sqrt{n+1}-1\right)\)
Vậy ta có điều phải chứng minh.
Mình cũng đang tìm câu hỏi như vậy. Ai biết làm giúp với
1.\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}=\frac{\left(5+\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\frac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\frac{25+10\sqrt{5}+5}{25-5}+\frac{25-10\sqrt{5}+5}{25-5}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{20}\)
\(=\frac{60}{20}=3\)
2.
a) \(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\)
ĐK : x ≥ 0
<=> \(\sqrt{5x\cdot9}-2\sqrt{5x\cdot4}+2\sqrt{5x\cdot16}=21\)
<=> \(\sqrt{5x\cdot3^2}-2\sqrt{2^2\cdot5x}+2\sqrt{5x\cdot4^2}=21\)
<=> \(\left|3\right|\sqrt{5x}-2\cdot\left|2\right|\sqrt{5x}+2\cdot\left|4\right|\sqrt{5x}=21\)
<=> \(\sqrt{5x}\cdot\left(3-4+8\right)=21\)
<=> \(\sqrt{5x}\cdot7=21\)
<=> \(\sqrt{5x}=3\)
<=> \(5x=9\)
<=> \(x=\frac{9}{5}\left(tm\right)\)
ơ đang làm lại bấm " Gửi trả lời " ._.
2b) \(\sqrt{x^2-10x+25}=4\)
<=> \(\sqrt{\left(x-5\right)^2}=4\)
<=> \(\left|x-5\right|=4\)
<=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
3. \(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
ĐK : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x-1}\right)}\right)\div\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
Bài 1:
a) 4x-\(\sqrt{9x^2-12x+4}\)
= 4x-\(\sqrt{\left(3x-2\right)^2}\)
= 4x-\(|3x-2|\)
= 4x-3x+2
= x+2
b) Thay x=\(\dfrac{2}{7}\)vào biểu thức A, ta có:
A= \(\dfrac{2}{7}+\dfrac{1}{2}\)= \(\dfrac{11}{14}\)
Bài 2:
a) \(\sqrt{x^2+2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow\)\(\left(\sqrt{x^2+2x+1}\right)^2=\left(\sqrt{x+1}\right)^2\)
\(\Leftrightarrow\)x2+2x+1=x+1
\(\Leftrightarrow\)x2+2x+1-x-1=0
\(\Leftrightarrow\)x2-x=0
\(\Leftrightarrow\)x(x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Bạn ơi, cả hai biểu thức này có ẩn là x chứ đâu có a mà bạn lại ghi là a>0 ???
Giải:
Đặt \(y=b\sqrt{1-x}\)
Ta có: \(\sqrt{a+y}=1+\sqrt{a-y}\)
\(\Leftrightarrow\sqrt{a+y}-\sqrt{a-y}=1\)
\(\Leftrightarrow\left(\sqrt{a+y}-\sqrt{a-y}\right)^2=1\)
\(\Leftrightarrow a+y-2\cdot\sqrt{a+y}\cdot\sqrt{a-y}+a-y=1\)
\(\Leftrightarrow2a-2\sqrt{a^2-y^2}=1\)
\(\Leftrightarrow2\sqrt{a^2-y^2}=2a-1\)
\(\Leftrightarrow\sqrt{a^2-y^2}=\dfrac{2\left(a-\dfrac{1}{2}\right)}{2}=a-\dfrac{1}{2}\)
\(\Leftrightarrow a^2-y^2=\left(a-\dfrac{1}{2}\right)^2=a^2-a+\dfrac{1}{4}\)
\(\Leftrightarrow y^2=a-\dfrac{1}{4}\)
\(\Leftrightarrow y=\sqrt{a-\dfrac{1}{4}}\)
\(\Leftrightarrow b\sqrt{1-x}=\sqrt{a-\dfrac{1}{4}}\)
\(\Leftrightarrow\sqrt{1-x}=\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\)
\(\Leftrightarrow1-x=\left(\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\right)^2\)
\(\Leftrightarrow x=1-\left(\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\right)^2\)
Vậy....................