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a) \(\sqrt{\left(\sqrt{7-2}\right)^2}=\sqrt{5}\)
b)\(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-3\sqrt{2}\right)^2}\)
=\(\sqrt{2}-1-2+3\sqrt{2}=4\sqrt{2}-3\)
c)\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=2\sqrt{3}\)
d) hình như bn ghi sai
e)\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}\right):\sqrt{2}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{3}+1}\right):\sqrt{2}\)
=\(\dfrac{\sqrt{2+\sqrt{3}}\left(\sqrt{3}+1\right)+\sqrt{2-\sqrt{3}}\left(\sqrt{3}-1\right)}{2\sqrt{2}}\)
=\(\dfrac{\sqrt{6+3}+\sqrt{2+\sqrt{3}}+\sqrt{6-3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{2+\sqrt{3}}+\sqrt{3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{3}}{2\sqrt{2}}\)
f) \(\sqrt{9a^2}+3a-7=-3a+3a-7=-7\)
g)\(\dfrac{\sqrt{4x^2-4x+1}}{4x-2}+3x+2\)
=\(\dfrac{\sqrt{\left(2x-1\right)^2}}{4x-2}+3x+2=\dfrac{2x-1}{2\left(2x-1\right)}+3x+2\)
=\(\dfrac{1}{2}+3x+2=\dfrac{5}{2}+3x\)
h)\(\sqrt{\left(5a-1\right)^2}+2a-3\)
nếu a<0 :\(-5a+1+2a-3=-3a-2\)
nếu a>0 : \(5a-1+2a-3=7a-4\)
i)\(\sqrt{\dfrac{2a}{5}}.\sqrt{\dfrac{5a}{18}}+2\left(a-1\right)\)
=\(\sqrt{\dfrac{10a^2}{90}}+2a-2=\sqrt{\dfrac{a^2}{9}}+2a-2\)
=\(\dfrac{a}{3}+2a-2=\dfrac{7a}{3}-2\)
a) \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=-10\sqrt{2}+5.2-\left(18-30\sqrt{2}+25\right)\)
\(=-10\sqrt{2}+10-18+30\sqrt{2}-25\)
\(=20\sqrt{2}-33\)
b) câu b đề sai
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
a, \(\sqrt{\left(2-\sqrt{5}\right)^2}=\sqrt{5}-2\left(\sqrt{5}>2\right)\)
b, \(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\left(3>\sqrt{2}\right)\)
c, Với a < 3
\(\sqrt{\left(a-3\right)^2}+\left(a-9\right)=3-a+a-9=-6\)
d, \(A=\sqrt{\left(2a+5\right)^2}-\left(2a-7\right)\)
\(=\left|2a+5\right|-2a+7\)
+) Xét \(x\ge\dfrac{-5}{2}\) có:
\(A=2a+5-2a+7=12\)
+) Xét \(x< \dfrac{-5}{2}\) có:
\(A=-2a-5-2a+7=-4a+2\)
Vậy...
a: \(A=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)
\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{-4\sqrt{a}}{a-1}\)
\(=\dfrac{-\left(a-1\right)}{\sqrt{a}}\)
b: \(=1+\left(\dfrac{\left(2\sqrt{a}-1\right)}{1-\sqrt{a}}+\dfrac{2a\sqrt{a}-\sqrt{a}+a}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\)
Δ\(=1+\left(\dfrac{\left(-2\sqrt{a}+1\right)}{\sqrt{a}-1}+\dfrac{2a\sqrt{a}-\sqrt{a}+a}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\left(\dfrac{-2a\sqrt{a}-\sqrt{a}+1+2a\sqrt{a}-\sqrt{a}+a}{a+\sqrt{a}+1}\cdot\dfrac{\sqrt{a}}{2\sqrt{a}-1}\right)\)
\(=1+\dfrac{\left(\sqrt{a}-1\right)^2\cdot\sqrt{a}}{\left(2\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{2a\sqrt{a}+2a+2\sqrt{a}-a-\sqrt{a}-1+a\sqrt{a}-2a+\sqrt{a}}{\left(2\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{3a\sqrt{a}-a+2\sqrt{a}-1}{\left(2\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
Câu 2:
a: \(=2\left(\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\cdot\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=8\)
b: \(=\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{a-1}\cdot\left(\dfrac{a+1-2}{a+1}\right)^2\)
\(=\dfrac{2\left(a+1\right)}{a-1}\cdot\dfrac{\left(a-1\right)^2}{\left(a+1\right)^2}=\dfrac{2\left(a-1\right)}{a+1}\)
Từ \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+2017\)
\(\Leftrightarrow7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2017\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le2017\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(T=\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2c^2+a^2\right)}}\)
\(\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2^2}{2a}+\dfrac{1^2}{b}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2b}+\dfrac{1^2}{c}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2c}+\dfrac{1^2}{a}\right)\)
\(\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\)\(=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\le\sqrt{\left(\dfrac{1}{81}+\dfrac{1}{81}+\dfrac{1}{81}\right)\left(\dfrac{9}{a^2}+\dfrac{9}{b^2}+\dfrac{9}{c^2}\right)}\)
\(\le\sqrt{\dfrac{1}{81}\cdot3\cdot9\cdot2017}=\sqrt{\dfrac{2017}{3}}\)
Vậy \(T_{Max}=\sqrt{\dfrac{2017}{3}}\) khi \(a=b=c=\sqrt{\dfrac{3}{2017}}\)
So kimochiii~
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
\(=\dfrac{4+a+4\sqrt{a}-a+6\sqrt{a}-9}{2a-\sqrt{a}}=\)
\(=\dfrac{10\sqrt{a}-5}{2a-\sqrt{a}}=\dfrac{5\left(2\sqrt{a}-1\right)}{\sqrt{a}\left(2\sqrt{a-1}\right)}=\)
\(=\dfrac{5\sqrt{a}}{a}\)
Lời giải:
ĐKXĐ: $a>0; a\neq \frac{1}{4}$
\(A=\frac{(2+\sqrt{a}-\sqrt{a}+3)(2+\sqrt{a}+\sqrt{a}-3)}{\sqrt{a}(2\sqrt{a}-1)}=\frac{5(2\sqrt{a}-1)}{\sqrt{a}(2\sqrt{a}-1)}=\frac{5}{\sqrt{a}}\)
P/s: Lần sau bạn lưu ý ghi đầy đủ yêu cầu đề bài