Toán 9 Bài tập đại số

matheverytime · 1 Tháng sáu 2018

caau 1) a)

A\leqslant \frac{\frac{a^2+b^2}{2}}{\sqrt{2(a^2+b^2)}+2}=\frac{2}{2\sqrt{2}+2}=\frac{1}{\sqrt{2}+1}

caau b)

VP=\frac{x^3+y^3+z^3}{2xyz}=\frac{x^2}{2yz}+\frac{y^2}{2xz}+\frac{z^2}{2xy}+3

VT=\frac{x^2+y^2+z^2}{x^2+y^2}+\frac{x^2+y^2+z^2}{z^2+y^2}+\frac{x^2+y^2+z^2}{x^2+z^2}=3+\frac{z^2}{x^2+y^2}+\frac{y^2}{x^2+z^2}+\frac{x^2}{z^2+y^2}\leqslant 3+\frac{z^2}{2xy}+\frac{x^2}{2zy}+\frac{y^2}{2xz}=VP

cau 2)a)

x^2+6x+9+3x+10-2\sqrt{3x+10}+1=0<=>(x+3)^2+(\sqrt{3x+10}-1)^2=0=>x=-3

b)

2x^2+2(-x^2y^2-y^2)+3=-y^3<=>2x^2(1-y^2)-2(y^2-1)=-y^3-1<=>(1+y)[2x^2(1-y)-2(y-1)+y^2-y+1]=>y=1\vee 2x^2(1-y)+y^2-3y+3=0

2x^2(1-y)+y^2-3y+3=0

( vô lý vì )

2(x-1)^2+1=-y^3\geqslant 1 =>y\leqslant -1

=>y=-1=>x=1
câu 3)

a^2=x^2+y^2+\sqrt[3]{x^2y^2}(\sqrt[3]{x^2}+\sqrt[3]{y^2})+2\sqrt{2x^2y^2+x^2\sqrt[3]{x^2y^4}+y^2\sqrt[3]{x^4y^2}}=x^2+y^2+\sqrt[3]{x^2y^2}(\sqrt[3]{x^2}+\sqrt[3]{y^2})+2\sqrt[3]{x^2y^2}(\sqrt[3]{x^2}+\sqrt[3]{y^2})=(\sqrt[3]{x^2}+\sqrt[3]{y^2})^3=>\sqrt[3]{a^2}=\sqrt[3]{x^2}+\sqrt[3]{y^2}