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\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
a: =(2căn 3-8căn 3)(căn 3-1)
=-6căn 3*(căn 3-1)
=-18+6căn 3
b: \(=\dfrac{6-2\sqrt{5}}{\sqrt{5}-3}-\sqrt{5}+2\)
=-2-căn 5+2=-căn 5
c: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)
=\(3\sqrt{2a}-3a\cdot\sqrt{2a}\)
\(a,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ b,=2a-6b+6b-5a=-3a\)
Xí câu BĐT:
ta cần chứng minh \(\dfrac{a^2}{b^2c}+\dfrac{b^2}{c^2a}+\dfrac{c^2}{a^2b}\ge\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)
Áp dụng BĐT cauchy:
\(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}.ab}=2a^2\)
tương tự ta có:\(\dfrac{b^3}{c}+bc\ge2b^2;\dfrac{c^3}{a}+ac\ge2c^2\)
cả 2 vế các BĐT đều dương,cộng vế với vế ta có:
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2a^2+2b^2+2c^2\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\)
mà a2+b2+c2\(\ge ab+bc+ca\) ( chứng minh đầy đủ nhá)
do đó \(S=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(ab+bc+ca\right)-ab+bc+ca=ab+bc+ca\)
suy ra BĐT ban đầu đúng
dấu = xảy ra khi và chỉ khi a=b=c.
P/s: cách khác :Áp dụng BĐT cauchy-schwarz:
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)
\(S\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)
Câu hệ này =))
b, Từ hệ đã cho ta thấy x,y > 0
Trừ vế cho vế pt (1) và (2) của hệ ta được:
\(x^4-y^4=4y-4x\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)=4\left(y-x\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)+4\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(x+y\right)\left(x^2+y^2\right)+4\right]=0\)
\(\Leftrightarrow x-y=0\) ( Vì \(\left(x+y\right)\left(x^2+y^2\right)+4>0\) với x,y > 0)
\(\Leftrightarrow x=y\)
Với x = y thay vào pt đầu của hệ ta được:
\(x^4-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow x-1=0\) ( Vì \(x^2+2x+3>0\) )
\(\Leftrightarrow x=1\)
Với x=1 suy ra y=1
Vậy hệ đã cho có nghiệm duy nhất (x;y) = (1;1)
\(a,=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ =\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\\ b,=\dfrac{x^2+4x+3+x^2+4x+4}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{x+1}{x+3}\left(x\ne-1;x\ne-2;x\ne-3\right)\\ =\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)
\(a,\dfrac{4}{\sqrt{x}+1}+\dfrac{2}{1-\sqrt{x}}-\dfrac{\sqrt{x}-5}{x-1}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
\(b,\left(\dfrac{x+1}{x+2}+\dfrac{x+2}{x+3}\right):\dfrac{x+3}{x+1}\)
\(=\left(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x+3\right)}\right).\dfrac{x+1}{x+3}\)
\(=\left(\dfrac{x^2+4x+3}{\left(x+2\right)\left(x+3\right)}+\dfrac{x^2+4x+4}{\left(x+2\right)\left(x+3\right)}\right).\dfrac{x+1}{x+3}\)
\(=\dfrac{x^2+4x+3+x^2+4x+4}{\left(x+2\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)
\(=\dfrac{2x^2+8x+7}{\left(x+2\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)
\(=\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{\left(2x^2+8x+7\right).x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{2x^3+8x^2+7x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{2x^3+10x^2+15x+7}{\left(x+2\right)\left(x+3\right)^2}\)
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
?
\(A=\dfrac{x^3}{9y^2}-\dfrac{1}{8}x^2y+\dfrac{2}{15}xy^2\\ B=\dfrac{2a-b}{a+1}-\dfrac{\left(a-1\right)^2}{b-2}\cdot\dfrac{\left(b-2\right)\left(b+2\right)}{\left(a-1\right)\left(a+1\right)}\\ B=\dfrac{2a-b}{a+1}-\dfrac{\left(a-1\right)\left(b+2\right)}{a+1}\\ B=\dfrac{2a-b-\left(a-1\right)\left(b+2\right)}{a+1}\\ B=\dfrac{2a-b-ab-2a+b+2}{a+1}=\dfrac{2-ab}{a+1}\)