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\(a, A=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=\left(2-3-4\right)\sqrt{x-1}=-5\sqrt{x-1}\)
\(b, B=\frac{2}{x+y}.\left(x+y\right)\sqrt{\frac{3}{4}}=2\sqrt{\frac{3}{4}}=2.\frac{1}{2}.\sqrt{3}=\sqrt{3}\)
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
a. =\(\frac{x\sqrt{xy}+y\sqrt{x^2}-x\sqrt{y^2}-y\sqrt{xy}}{\sqrt{xy}}\)=\(\frac{x\sqrt{xy}+xy-xy-y\sqrt{xy}}{\sqrt{xy}}\)
=\(\frac{x\sqrt{xy}-y\sqrt{xy}}{\sqrt{xy}}\)=\(\frac{\sqrt{xy}\left(x-y\right)}{\sqrt{xy}}\)=\(x-y\)
b. =\(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x-1}}\)=\(x+\sqrt{x}+1\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
1)
a) Ta có : \(\frac{x^2+5}{\sqrt{x^2+4}}=\frac{\left(x^2+4\right)+1}{\sqrt{x^2+4}}=\sqrt{x^2+4}+\frac{1}{\sqrt{x^2+4}}\). Đến đây áp dụng bđt \(a+\frac{1}{a}>2\)là ra nhé :)
b) Ta sẽ chứng minh bằng biến đổi tương đương :
\(\sqrt{\left(a+c\right)\left(b+d\right)}\ge\sqrt{ab}+\sqrt{cd}\)
\(\Leftrightarrow\left(a+c\right)\left(b+d\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)
\(\Leftrightarrow ab+ad+bc+cd\ge ab+cd+2\sqrt{abcd}\)
\(\Leftrightarrow ad-2\sqrt{abcd}+bc\ge0\)
\(\Leftrightarrow\left(\sqrt{ad}-\sqrt{bc}\right)^2\ge0\)(luôn đúng)
Vì bđt cuối luôn đúng nên bđt ban đầu được chứng minh.
2) Mình làm tóm tắt thôi nhé , do đề dài...
a) \(\sqrt{2x+\sqrt{4x-1}}-\sqrt{2x-\sqrt{4x-1}}\)
\(=\frac{\sqrt{\left(4x-1\right)+2\sqrt{4x-1}+1}+\sqrt{\left(4x-1\right)-2\sqrt{4x-1}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}}{\sqrt{2}}=\frac{\left|\sqrt{4x-1}-1\right|+\left|\sqrt{4x-1}+1\right|}{\sqrt{2}}\)
b) \(\frac{x-y+3\sqrt{x}+3\sqrt{y}}{\sqrt{x}-\sqrt{y}+3}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+3\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}+3}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+3\right)}{\sqrt{x}-\sqrt{y}+3}=\sqrt{x}+\sqrt{y}\)
c) Biến đổi : \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}-1\right|\)
d) Biến đổi tương tự c)
e) \(\sqrt{x+\sqrt{x^2-4}}.\sqrt{x-\sqrt{x^2-4}}=\sqrt{x^2-\left(x^2-4\right)}=\sqrt{4}=2\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
a,\(\sqrt{x-4+4\sqrt{x-4}+4}\) +\(\sqrt{x-4-4\sqrt{x-4}+4}\)
=\(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\) (vi x>=8)
=\(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
b, \(\sqrt{x-1+2\sqrt{x\left(x-1\right)}+x}+\sqrt{x-1-2\sqrt{x\left(x-1\right)}+x}\)
=\(\sqrt{x-1}+\sqrt{x}+\left|\sqrt{x-1}-\sqrt{x}\right|\)
=\(\sqrt{x}+\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}\) =\(2\sqrt{x}\)
c,d sai dau bai hay sao y