0
phiếu
0đáp án
1 lượt xem

.

C/m: $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq9$
2
phiếu
1đáp án
567 lượt xem

toan 8

cm bieu thc sau ko am vs moi x$(x^2+1)^4+9(x^2+1)^3+21(x^2+1)^3+(x^2+1)-29 $
2
phiếu
1đáp án
740 lượt xem

Toán

Rút gọn$M=1+\frac{x+3}{x^{2}+5x+6}\div (\frac{8x}{4x^{3}-8x^{2}}-\frac{3x}{3x^{2}-12}-\frac{1}{x+2})$
3
phiếu
1đáp án
880 lượt xem

hô hô mấy mem HTn đâu rồi ra đây xử lý giùm bt này

Cho : x, y, z là các số thực ko âm. CMR : $3(x^2+y^2+z^2)\geq (x+y+z)(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})+(x-y)^2+(y-z)^2+(z-x)^2\geq (x+y+z)^2$
3
phiếu
1đáp án
883 lượt xem

Chiuu, Jin , Ngốc và cả thảy mem HTN tiếp chiêu bất đẳng bộ pháp đây. ( chú thích: giải hậu tạ )

cho các số dương a,b,c,d . Chứng minh rằng: a2b5+b2c5+c2d5+d2a5≥1a3+1b3+1c3+1d3" role="presentation" style="display: inline-block;...
1
phiếu
0đáp án
454 lượt xem
4
phiếu
2đáp án
2K lượt xem

BĐT hay va kho

cho $x,y,z > 0$ tìm giá trị nhỏ nhất?$P=\sqrt[3]{4(x^{3}+y^{3})}+\sqrt[3]{4(x^{3}+z^{3})}+\sqrt[3]{4(z^{3}+y^{3})}+2(\frac{x}{y^{2}}+\frac{y}{z^{2}}+\frac{z}{x^{2}})$

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