Chứng minh rằng :
a) \(i+i^2+i^3+...+i^{99}+i^{100}=0\)
b) \(\dfrac{\left(\sqrt{2}+i\right)\left(1-i\right)\left(1+i\right)}{i}=2-2\sqrt{2}i\)
Những câu hỏi liên quan
Thực hiện các phép tính :
a) \(\left(2+3i\right)\left(3-i\right)+\left(2-3i\right)\left(3+i\right)\)
b) \(\dfrac{2+i\sqrt{2}}{1-i\sqrt{2}}+\dfrac{1+i\sqrt{2}}{2-i\sqrt{2}}\)
c) \(\dfrac{\left(1+i\right)\left(2+i\right)}{2-i}+\dfrac{\left(1+i\right)\left(2-i\right)}{2+i}\)
Rút gọn biểu thức :
\(\frac{\sqrt{7-4\sqrt{3}}}{\sqrt{2-\sqrt{3}}}.\sqrt{2+\sqrt{3}}\)
\(\left[\left(a-b\right)\sqrt{\frac{a+b}{a-b}}+a-b\right]\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)với a>b>0
Chứng minh rằng :
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
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Câu 1: Cho bt: A left(dfrac{1}{sqrt{1+x}}+sqrt{1-x}right):left(dfrac{1}{sqrt{1-x^2}}+1right)
a) Tìm x để A có nghĩa
b) Rút gọn
c) Tính A với x dfrac{sqrt{3}}{2+sqrt{3}}
Câu 2: Cho bt B left(dfrac{x-y}{sqrt{x}-sqrt{y}}+dfrac{xsqrt{x}-ysqrt{y}}{y-x}right):left(dfrac{left(sqrt{x}-sqrt{y}right)^2+sqrt{xy}}{sqrt{x}+sqrt{y}}right)
a) Rút gọn
b) CM Bge0
c) So sánh B với sqrt{B}
Đọc tiếp
Câu 1: Cho bt: A= \(\left(\dfrac{1}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{1}{\sqrt{1-x^2}}+1\right)\)
a) Tìm x để A có nghĩa
b) Rút gọn
c) Tính A với x =\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\)
Câu 2: Cho bt B= \(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\)
a) Rút gọn
b) CM B\(\ge\)0
c) So sánh B với \(\sqrt{B}\)
Bài 1. Tìm x, y, z biết: \(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\) (trong đó, a + b + c = 3)
Bài 2.
a) Chứng minh rằng: \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b/ Cho S = \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\). Chứng minh rằng: 18<S<19
Câu 1:
\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\\ \Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}=0\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}+3-a-b-c=0\\ \Leftrightarrow\left[\left(x-a\right)-2\sqrt{x-a}+1\right]+\left[\left(y-b\right)-2\sqrt{y-b}+1\right]+\left[\left(z-c\right)-2\sqrt{z-c}+1\right]=0\\ \Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}-1=0\\\sqrt{y-b}-1=0\\\sqrt{z-c}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}=1\\\sqrt{y-b}=1\\\sqrt{z-c}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-a=1\\y-b=1\\z-c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=a+1\\y=b+1\\z=c+1\end{matrix}\right.\)Vậy \(\left\{x;y;z\right\}=\left\{a+1;b+1;c+1\right\}\)
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Câu 2:
\(\text{ a) Ta có }:\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\\ =\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\left(1\right)\)
\(\text{Lại có: }\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}-\sqrt{n}\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow2\left(\sqrt{n+1}-n\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b) Áp dụng bất đảng thức ở câu a:
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\\ >2\left(\sqrt{101}-\sqrt{100}\right)+...+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{100}+...+\sqrt{4}-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-1\right)=2\left(10-1\right)=18\left(3\right)\)
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 2\left(\sqrt{100}-\sqrt{99}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-\sqrt{0}\right)\\ =2\left(\sqrt{100}-\sqrt{99}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\sqrt{1}\right)\\ =2\cdot\sqrt{100}=2\cdot10=20\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\Rightarrow18< S< 20\)
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Xem thêm câu trả lời
1. Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}\left(a+b\right)^2=10+2ab\\\left(a+b\right)\left(a-\dfrac{2}{ab}\right)=\dfrac{4}{3}\end{matrix}\right.\)
2.Giải phương trình:
\(\sqrt{\sqrt{2}-1-x}+\sqrt[4]{x}=\dfrac{1}{\sqrt[4]{2}}\)
Giải các phương trình sau trên tập số phức :
a) \(3x^2+\left(2+2i\sqrt{2}\right)x-\dfrac{\left(1+i\right)^3}{1-i}=i\sqrt{8}x\)
b) \(\left(1-ix\right)^2+\left(3+2i\right)x-5=0\)
Bài 1: Cho A dfrac{1}{2sqrt{1}+1sqrt{2}}+dfrac{1}{3sqrt{2}+2sqrt{3}}+dfrac{1}{4sqrt{3}+3sqrt{4}}+...+dfrac{1}{100sqrt{99}+99sqrt{100}}
So sánh A với 1
Bài 2: Tính
A left(dfrac{3}{sqrt{2}+1}+dfrac{14}{2sqrt{2}-1}-dfrac{4}{2-sqrt{2}}right).left(sqrt{8}+2right)
Bài 3: Tính tổng
Sdfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+dfrac{1}{sqrt{4}+sqrt{5}}+...+dfrac{1}{sqrt{2018}+sqrt{2019}}
Đọc tiếp
Bài 1: Cho A = \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
So sánh A với 1
Bài 2: Tính
A = \(\left(\dfrac{3}{\sqrt{2}+1}+\dfrac{14}{2\sqrt{2}-1}-\dfrac{4}{2-\sqrt{2}}\right).\left(\sqrt{8}+2\right)\)
Bài 3: Tính tổng
S=\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+...+\dfrac{1}{\sqrt{2018}+\sqrt{2019}}\)
Bài 1:Với mọi n∈N*,ta có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó :
A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
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Bài 2:
\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
=10
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Bài 1: Tính:
Aleft(-2right).left(-3right)-5.left|-5right|+125.left(dfrac{-1}{5}right)^2
Bleft(-3right).left|-7right|-left(-4right).left|5right|+dfrac{1}{3}.left|-9right|
Cleft(-2right)^3.left|-3right|-dfrac{1}{5}.left|-25right|-4.left|-7right|+left(-2right)^2
Dleft(-6right).left|-3right|+2.left|-9right|-7left|left(-2right)^3right|+8.left|-7right|
Eleft|-3^2right|.left|4right|-left|7right|.8-left|6right|.left|-8right|-left|12right|.left(dfrac{1}{2}right)^2
Bài 2: Tìm x:
a)12-2left|3x+2rig...
Đọc tiếp
Bài 1: Tính:
A=\(\left(-2\right).\left(-3\right)-5.\left|-5\right|+125.\left(\dfrac{-1}{5}\right)^2\)
B=\(\left(-3\right).\left|-7\right|-\left(-4\right).\left|5\right|+\dfrac{1}{3}.\left|-9\right|\)
C=\(\left(-2\right)^3.\left|-3\right|-\dfrac{1}{5}.\left|-25\right|-4.\left|-7\right|+\left(-2\right)^2\)
D=\(\left(-6\right).\left|-3\right|+2.\left|-9\right|-7\left|\left(-2\right)^3\right|+8.\left|-7\right|\)
E=\(\left|-3^2\right|.\left|4\right|-\left|7\right|.8-\left|6\right|.\left|-8\right|-\left|12\right|.\left(\dfrac{1}{2}\right)^2\)
Bài 2: Tìm x:
a)\(12-2\left|3x+2\right|=10\)
b)\(2.\left|5-4x\right|+17=\left(-2\right)^3.\left(-4\right)\)
c)\(\left|3x-5\right|+\left(-3\right)^2.2=12.\left|3x+5\right|+117\)
d)\(4.\left|3-2x\right|+\left(-5\right).\left|4-3x\right|-5=-6\)
e)\(\left|2x-7\right|-2^3.\left|2x-7\right|+15=-5.\left|2x-7\right|+3\)
f)\(\left|x+2\right|+\left|x^2-4\right|=0\)
g)\(\left|3x-9\right|+\left|x^2-9\right|=0\)
h)\(\left|2x-1\right|+\left|x^2-\dfrac{1}{4}\right|=0\)
1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
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1. B = (-3).|-7| - (-4).|5| + \(\dfrac{1}{3}.\left|-9\right|\)
= -21 + 20 + 3
= 2
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Xem thêm câu trả lời
Mình đang cần gấp giải giùm mình với, mình sẽ tặng 1SP
1, chứng minh đẳng thức
d, left(1+dfrac{a+sqrt{a}}{sqrt{a}+1}right)left(1-dfrac{a-sqrt{a}}{sqrt{a}-1}right)1-a
2, cho biểu thức
Qdfrac{a}{sqrt{a^2-b^2}}-left(1+dfrac{a}{sqrt{a^2-b^2}}right):dfrac{b}{a-sqrt{a^2-b^2}} với ab0
a) Rút gọn Q
b0 Xác địng giá trị của Q khi a 3b
Đọc tiếp
Mình đang cần gấp 
giải giùm mình với, mình sẽ tặng 1SP 
1, chứng minh đẳng thức
d, \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)
2, cho biểu thức
Q=\(\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\) với a>b>0
a) Rút gọn Q
b0 Xác địng giá trị của Q khi a = 3b
1. \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
2. a) Với a>b>0 thì
\(Q=\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\)
\(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\dfrac{a-\sqrt{a^2-b^2}}{b}\)
\(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{b^2}{b\sqrt{a^2-b^2}}=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{b}{\sqrt{a^2-b^2}}\)
\(=\dfrac{a-b}{\sqrt{a^2-b^2}}=\dfrac{a-b}{\sqrt{a-b}.\sqrt{a+b}}=\sqrt{\dfrac{a-b}{a+b}}\)
b) Thay a = 3b ta được
\(Q=\sqrt{\dfrac{a-b}{a+b}}=\sqrt{\dfrac{3b-b}{3b+b}}=\sqrt{\dfrac{2b}{4b}}=\sqrt{\dfrac{1}{2}}=\dfrac{\sqrt{2}}{2}\)
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1) d) ta có : \(VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)
\(\Rightarrow\) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\) (đpcm)
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1. \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)
vt=\(\left(1+\sqrt{a}\right).\left(1-\sqrt{a}\right)=1-a=vp\left(đpcm\right)\)
2. a. Q=\(\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\left(a>b>0\right)\)
=\(\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{a^2-\left(a^2-b^2\right)}{b.\sqrt{a^2-b^2}}\)
=\(\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{b^2}{b\sqrt{a^2-b^2}}\)
=\(\dfrac{a-b}{\sqrt{a^2-b^2}}=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}\)
b. Với a=3b thay vào biểu thức đã cho, ta được:
\(Q=\dfrac{\sqrt{3b-b}}{\sqrt{3b+b}}=\dfrac{\sqrt{2b}}{\sqrt{4b}}=\dfrac{\sqrt{2}}{2}\)
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