giúp mình bài chứng minh với ><
\(c\text{hứng minh \text{rằng :}}\dfrac{4}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}-\sqrt{\sqrt{5}-2}=1\)
Bài 2.4 Chứng minh với mọi giá trị của x để biểu thức có nghĩa thì giá trị của:
𝐴= \((\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x+3}}{2\sqrt{x}+2}).\dfrac{4x-4}{5}\) không phụ thuộc vào x.
ai bt giúp mình với mình đang cần gấp
A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)
A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)
A = 4
Vậy A không phụ thuộc vào x
Chúc bn học tốt!
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)
\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)
\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)
\(=10\cdot\dfrac{2}{5}=4\)
Bài 1 :
a) Cho 3 số hữu tỉ a,b,c thoả mãn : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\). Chứng minh rằng : \(A\text{=}\sqrt{a^2+b^2+c^2}\) là số hữu tỉ.
b) Cho 3 số x,y,z đôi một khác nhau . Chứng minh rằng : \(B\text{=}\sqrt{\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y-z\right)^2}+\dfrac{1}{\left(z-x\right)^2}}\) là một số hữu tỉ.
a) Từ giả thiết : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)
\(\Rightarrow2ab\text{=}2bc+2ca\)
\(\Rightarrow2ab-2bc-2ca\text{=}0\)
Ta xét : \(\left(a+b-c\right)^2\text{=}a^2+b^2+c^2+2ab-2bc-2ca\)
\(\text{=}a^2+b^2+c^2\)
Do đó : \(A\text{=}\sqrt{a^2+b^2+c^2}\text{=}\sqrt{\left(a+b-c\right)^2}\)
\(\Rightarrow A\text{=}a+b-c\)
Vì a;b;c là các số hữu tỉ suy ra : đpcm
b) Đặt : \(a\text{=}\dfrac{1}{x-y};b\text{=}\dfrac{1}{y-x};c\text{=}\dfrac{1}{z-x}\)
Do đó : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)
Ta có : \(B\text{=}\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)
Từ đây ta thấy giống phần a nên :
\(B\text{=}a+b-c\)
\(B\text{=}\dfrac{1}{x-y}+\dfrac{1}{y-z}-\dfrac{1}{z-x}\)
Suy ra : đpcm.
Mình bổ sung đề phần b cần phải có điều kiện của x;y;z nha bạn.
Cho a= \(\sqrt{3\text{+}\sqrt{5\text{+}2\sqrt{3}}}\) + \(\sqrt{3-\sqrt{5\text{+}2\sqrt{3}}}\)
Chứng minh rằng a\(^2\) - 2a - 2 = 0
Lời giải:
Ta có:
$a^2=3+\sqrt{5+2\sqrt{3}}+3-\sqrt{5+2\sqrt{3}}+2\sqrt{(3+\sqrt{5+2\sqrt{3}})(3-\sqrt{5+2\sqrt{3}})}$
$=6+2\sqrt{3^2-(5+2\sqrt{3})}=6+2\sqrt{4-2\sqrt{3}}=6+2\sqrt{3+1-2\sqrt{3}}$
$=6+2\sqrt{(\sqrt{3}-1)^2}=6+2(\sqrt{3}-1)=4+2\sqrt{3}=(\sqrt{3}+1)^2$
$\Rightarrow a=\sqrt{3}+1$ (do $a\geq 0$)
Do đó:
$a^2-2a-2=4+2\sqrt{3}-2(\sqrt{3}+1)-2=0$ (đpcm)
1. CHỨNG MINH ĐẲNG THỨC
a. \(\text{[}3+2\sqrt{6}-\sqrt{33}\text{]}\cdot\text{[}\sqrt{22}+\sqrt{6}+4\text{]}=24\)
b. \(\text{[}\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\text{]}\cdot\text{[}15+2\sqrt{6}\text{]}\)
c.\(\text{[}\frac{4}{3}\cdot\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\text{]}\cdot\text{[}\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\text{]}=4\)
d. \(\sqrt{\text{[}1-\sqrt{1989}\text{]}^2}\cdot\sqrt{1990+2\sqrt{1989}}=1988\)
e. \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)với \(a>0;b>0\)và \(a\ne b\)
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
bài 1; a)Viết biểu thức trong căn dưới dạng tích rồi tính
\(\sqrt{117^2-108^2}\)
b) Chứng minh: \(\text{}\sqrt{9-4\sqrt{5}}+2=\sqrt{5}\)
a) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9\cdot225}=\sqrt{3^2\cdot15^2}=\left|3\cdot15\right|=45\)
b) \(\sqrt{9-4\sqrt{5}}+2=\sqrt{5-4\sqrt{5}+4}+2=\sqrt{\left(\sqrt{5}-2\right)^2}+2=\left|\sqrt{5}-2\right|+2=\sqrt{5}\)
\(a,\sqrt{117^2-108^2}\\ =\sqrt{\left(117-108\right)\left(117+108\right)}\\ =\sqrt{9.225}\\ =\sqrt{3^2}.\sqrt{15^2}\\ =3.15\\ =45\)
\(b,\sqrt{9-4\sqrt{5}}+2=\sqrt{5}\)
\(VT=\sqrt{9-4\sqrt{5}}+2\\ =\sqrt{\sqrt{5^2}-2.2\sqrt{5}+2^2}+2\\ =\sqrt{\left(\sqrt{5}-2\right)^2}+2\\ =\left|\sqrt{5}-2\right|+2\\ =\sqrt{5}-2+2\\ =\sqrt{5}=VP\left(dpcm\right)\)
a) \(\sqrt{117^2-108^2}\)
\(=\sqrt{\left(117-108\right)\left(117+108\right)}\)
\(=\sqrt{9\cdot225}\)
\(=\sqrt{3^2\cdot15^2}\)
\(=\sqrt{\left(3\cdot15\right)^2}\)
\(=3\cdot15\)
\(=45\)
b) VT: \(\sqrt{9-4\sqrt{5}}+2\)
\(=\sqrt{4-4\sqrt{5}+5}+2\)
\(=\sqrt{2^2-2\cdot2\cdot\sqrt{5}\cdot\left(\sqrt{5}\right)^2}+2\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|+2\)
\(=\sqrt{5}-2+2=\sqrt{5}=VP\) (đpcm)
Cho 3 số dương a;b;c thoả mãn : \(\sqrt{a^2+b^2}\text{+}\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\text{=}\sqrt{2011}\)
Chứng minh rằng : \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
* Chứng minh đẳng thức
\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=2\sqrt{x-1}\) với x ≥ 2
* Trục căn thức ở mẫu
a.\(\dfrac{1}{\sqrt{5}+\sqrt{7}}\)
b.\(\dfrac{2}{5-\sqrt{2}-\sqrt{3}}\)
c.\(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}\)
\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1+1}}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)
a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)
c) \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}=\dfrac{7}{2\sqrt{5}-\sqrt{3}}=\dfrac{7\left(2\sqrt{5}+\sqrt{3}\right)}{\left(2\sqrt{5}+\sqrt{3}\right)\left(2\sqrt{5}-\sqrt{3}\right)}\)
\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)
Chứng minh rằng với mọi số nguyên dương n ta đều có \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+\dfrac{1}{5\sqrt{4}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{(n+1)\sqrt{n}}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(n+1)\sqrt{n}}<\frac{(\sqrt{n+1}-\sqrt{n}).2\sqrt{n+1}}{(n+1)\sqrt{n}}\)
Hay \(\frac{1}{(n+1)\sqrt{n}}< \frac{2\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
Áp dụng vào bài toán:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{(n+1)\sqrt{n}}< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}+....+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}=2-\frac{2}{\sqrt{n+1}}< 2\)
Ta có đpcm.
giúp mình với nhanh hộ với
Chứng minh rằng
\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1\)
Đặt \(x=a+b=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(\Leftrightarrow x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(\Leftrightarrow x^3=2+\sqrt{5}+2-\sqrt{5}+\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}.x\)
\(\Leftrightarrow x^3=4+\sqrt[3]{4-5}.x\)
\(\Leftrightarrow x^3=4-3x\)
\(\Leftrightarrow x^3+3x-4=0\)
\(\Leftrightarrow x^3-x^2+x^2-x+4x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+4\right)=0\)
Vì \(x^2+x+4=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\left(\forall x\right)\)
Nên \(x-1=0\Leftrightarrow x=1\)
Vậy \(x=a+b=1\Rightarrow\) \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1\) (đpcm)
Dòng thứ 3 và thứ 4 bạn thiếu số 3 nhé @ Ngân@
\(3\sqrt[3]{\left(2+\sqrt{5}\right).\left(2-\sqrt{5}\right)}.x\)
Xin cảm ơn cô Linh Chi và bạn Lam Hàng ạ