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\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\4x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+4y=4\\4x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)
Ta có: \(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}=\sqrt{z\left[4\left(4-y-z\right)+yz\right]}\)
\(=\sqrt{x\left[4\left(x+\sqrt{xyz}\right)+yz\right]}=\sqrt{4x^2+4x\sqrt{xyz}+xyz}=2x+\sqrt{xyz}\)
Tương tự ta có: \(\sqrt{y\left(4-z\right)\left(4-z\right)}=2y+\sqrt{xyz}\)
Và: \(\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\)
Từ trên:
\(\Rightarrow T=2x+\sqrt{xyz}+2y+\sqrt{xyz}+2z+\sqrt{xyz}-\sqrt{xyz}\)
\(=2\left(x+y+z+\sqrt{xyz}\right)\)
\(=8\)
Thiếu đề rồi bạn ơi, bổ sung nhé:
Cho \(x+y+z+\sqrt{xyz}=4\)
Tính: \(M=\sqrt{x\left(4-y\right)\left(4-z\right)}+\sqrt{y\left(4-z\right)\left(4-x\right)}+\sqrt{z\left(4-x\right)\left(4-y\right)}-\sqrt{xyz}\)
Bài làm:
Ta có: \(\sqrt{x\left(4-x\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}\)
\(=\sqrt{x\left[4\left(4-y-z\right)+yz\right]}=\sqrt{x\left[4\left(x+\sqrt{xyz}\right)+yz\right]}\)
\(=\sqrt{x\left(4x+4\sqrt{xyz}+yz\right)}=\sqrt{x\left(2\sqrt{x}+\sqrt{yz}\right)^2}\)
\(=\sqrt{\left(2x+\sqrt{xyz}\right)^2}=2x+\sqrt{xyz}\)
Tương tự ta chứng minh được:
\(\sqrt{y\left(4-z\right)\left(4-x\right)}=2y+\sqrt{xyz}\) ; \(\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\)
=> \(M=2x+\sqrt{xyz}+2y+\sqrt{xyz}+2z+\sqrt{xyz}-\sqrt{xyz}\)
\(M=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)
\(x+y+z+\sqrt{xyz}=4\)
\(\Leftrightarrow xyz=\left(4-x-y-z\right)^2\)
\(\Leftrightarrow xyz=16+x^2+y^2+z^2-8x-8y-8z+2xy+2xz+yz\)
\(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}=\sqrt{16x-4xy-4xz+xyz}\)
\(=\sqrt{16x-4xy-4xz+16+x^2+y^2+z^2-8x-8y-8z+2xy+2yz+2xz}\)
\(=\sqrt{8x-2xy-2xz+2yz+x^2+y^2+z^2-8y-8z+16}\)
\(=\sqrt{\left(-x+y+z-4\right)^2}=\left|y+z-x-4\right|=\left|y+z-x-\left(x+y+z+\sqrt{xyz}\right)\right|\)
\(=\left|-2x-\sqrt{xyz}\right|=2x+\sqrt{xyz}\) (Vì x > 0)
Tương tự : \(\sqrt{y\left(4-z\right)\left(4-x\right)}=2y+\sqrt{xyz}\) , \(\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\)
Suy ra \(B=2x+2y+2z+2\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)
Ta có \(x+y+z+\sqrt{xyz}=4\Rightarrow4x+4y+4z+4\sqrt{xyz}=16\)
Ta lại có \(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}=\sqrt{x\left(4x+4\sqrt{xyz}+yz\right)}=\sqrt{4x^2+4x\sqrt{xyz}+xyz}=\sqrt{\left(2x+\sqrt{xyz}\right)^2}=2x+\sqrt{xyz}\)
Tương tự \(\sqrt{y\left(4-z\right)\left(4-x\right)}=2y+\sqrt{xyz}\)
\(\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\)
Suy ra \(P=\sqrt{x\left(4-y\right)\left(4-z\right)}+\sqrt{y\left(4-z\right)\left(4-x\right)}+\sqrt{z\left(4-x\right)\left(4-y\right)}-\sqrt{xyz}=2x+\sqrt{xyz}+2y+\sqrt{xyz}+2z+\sqrt{xyz}-\sqrt{xyz}=2x+2y+2z+2\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)
\(\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)
\(=\sqrt{5^2\left(y+4\right)}+\sqrt{6^2\left(y+4\right)}-2\sqrt{9^2\left(y+4\right)}\)
\(=5\sqrt{y+4}+6\sqrt{y+4}-2\cdot9\sqrt{y+4}\)
\(=11\sqrt{y+4}-18\sqrt{y+4}\)
\(=-7\sqrt{y+4}\)
\(=5\sqrt{y+4}+6\sqrt{y+4}-2\cdot9\sqrt{y+4}\)
\(=11\sqrt{y+4}-18\sqrt{y+4}\)
\(=-7\sqrt{y+4}\)