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a)\(\sqrt{x}=4\Leftrightarrow x=4^2\Leftrightarrow x=16\)
b)\(\sqrt{x-2}=3\Leftrightarrow x-2=3^2\Leftrightarrow x=9-2=7\)
c)\(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\Leftrightarrow\dfrac{x}{3}-\dfrac{7}{6}=\dfrac{1}{36}\Leftrightarrow\dfrac{x}{3}=-\dfrac{41}{36}\Leftrightarrow x=-\dfrac{41}{12}\)
d)\(x^2=7vớix< 0\)
\(\Leftrightarrow\left(-x\right)^2=7\Leftrightarrow-x=\sqrt{7}\Leftrightarrow x=-\sqrt{7}\)
e)\(x^2-4=0với>0\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\sqrt{4}=2\)
f)\(\left(2x+7\sqrt{7}\right)^2=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}+343=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}=-336\)
\(\Leftrightarrow4x^2=28\left(12-\sqrt{7}\right)\Leftrightarrow x^2=\dfrac{28\left(12-\sqrt{7}\right)}{4}=7\left(12-\sqrt{7}\right)\)
\(\Leftrightarrow x=\sqrt{7\left(12-\sqrt{7}\right)}=\sqrt{84-7\sqrt{7}}\)
Dạng tổng quát: \(\sqrt{a-b}\ge\sqrt{a}-\sqrt{b}\) với \(a\ge b\ge0\)
Chứng minh:
Ta có: \(\sqrt{a-b}\ge\sqrt{a}-\sqrt{b}\)
\(\Rightarrow\)\(\left(\sqrt{a-b}\right)^2\ge\left(\sqrt{a}-\sqrt{b}\right)^2\)
\(\Rightarrow\)\(a-b\ge a+b-2\sqrt{ab}\)
\(\Rightarrow\)\(-2b\ge-2\sqrt{ab}\)
\(\Rightarrow\)\(b\le\sqrt{ab}\)
\(\Rightarrow\)\(b^2\le ab\) luôn đúng do \(a\ge b\ge0\)
Vậy \(\sqrt{a-b}\ge\sqrt{a}-\sqrt{b}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b\)
Sau khi ib với Hoàng Nguyễn thì đề bài như sau
Tìm \(n\inℕ\)biết
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\)
ĐKXĐ: n > 1
Ta đi c/m bài toán tổng quát
\(\frac{1}{\sqrt{a-1}+\sqrt{a}}=\frac{\sqrt{a}-\sqrt{a-1}}{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}\)
\(=\frac{\sqrt{a}-\sqrt{a-1}}{a-a+1}\)
\(=\sqrt{a}-\sqrt{a-1}\)
Áp dụng vào bài toán đc
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\)
\(\Leftrightarrow\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n}-\sqrt{n-1}=11\)
\(\Leftrightarrow\sqrt{n-1}-1=11\)
\(\Leftrightarrow\sqrt{n-1}=12\)
\(\Leftrightarrow n-1=144\)
\(\Leftrightarrow n=145\left(TmĐKXĐ\right)\)
Vậy n = 145
a. \(\sqrt{35}+\sqrt{99}< \sqrt{36}+\sqrt{100}=6+10=16\)
\(\Rightarrow\sqrt{35}+\sqrt{99}< 16\)
b. \(\sqrt{24}< \sqrt{25}=5\)
\(\sqrt{5}+\sqrt{10}>\sqrt{4}+\sqrt{9}=2+3=5\)
\(\Rightarrow\sqrt{24}< \sqrt{5}+\sqrt{10}\)
\(\sqrt{2}+\sqrt{3}+\sqrt{5}< \sqrt{4}+\sqrt{9}+\sqrt{25}=2+3+5=10< 18\)
b) \(\sqrt{5}+\sqrt{7}+4< \sqrt{9}+\sqrt{9}+4=3+3+4=10< 12\)
a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)
=1+3+5+7+9
=25
b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)
=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)
=\(\dfrac{15}{12}\)
c) =0,2+0.3+0,4
= 0.9
d) =9-8+7
=8
j) =1,2-1,3+1.4
= (-0,1)+1,4
=1,4
g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)
= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)
= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)
=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)
= \(\dfrac{71}{20}\)
Nhớ tick cho mk nha~
a) Ta thấy số dưới lẫn số mũ của 536 lớn hơn 220 => 536>220
b)Ta có:\(99^{200}=99^{100}.99^{100}\)
\(9999^{100}=\left(99.101\right)^{100}=99^{100}.101^{100}\)
VÌ \(99^{100}.99^{100}< 99^{100}.101^{100}\)
Nên: \(99^{200}< 9999^{100}\)
c)Ta có: \(2^{150}=\left(2^3\right)^{50}=8^{50}\)
\(3^{100}=\left(3^2\right)^{50}=9^{50}\)
Vì \(8^{50}< 9^{50}\)nên : \(2^{150}< 3^{100}\)
d)\(\sqrt{26+2}=\sqrt{28}=5< x< 6\)
\(\sqrt{26}+\sqrt{2}=5< x< 6+1< y< 2\)
=> \(\sqrt{26+2}< \sqrt{26}+\sqrt{2}\)
Câu d mình l
\(=\dfrac{1}{2}.1,1-0,3+6=6,25\)