a,\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
b,\(\frac{27^2.8^5}{6^6.32^3}+\frac{3^4.4^4}{2^2.6^2}\)

c,\(\left(1+\frac{1}{1.3}\right)\)\(\left(1+\frac{1}{2.4}\right)\)\(\left(1+\frac{1}{3.5}\right)\)........\(\left(1+\frac{1}{20.22}\right)\)
d,\(\frac{3}{2}\)+\(\frac{7}{6}\)+\(\frac{13}{12}+\frac{21}{20}+.....+\frac{91}{90}\)
e,\(\left(-2^2\right)+\sqrt{36}-\sqrt{9}+\sqrt{25}\) 20).20)

\(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\)

\(VP=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\le\frac{a+b}{2}\sqrt{2\left(a+b\right)}\)\(\Rightarrow\)\(VP^2\le\frac{\left(a+b\right)^3}{2}\) (1) 

chứng minh bổ đề: \(VT^2=\left(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\right)^2\ge\frac{\left(a+b\right)^3}{2}\)

\(\Leftrightarrow\)\(\frac{\left(a+b\right)^4}{4}+\frac{\left(a+b\right)^2}{16}+\frac{\left(a+b\right)^3}{4}\ge\frac{\left(a+b\right)^3}{2}\)

\(\Leftrightarrow\)\(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge\left(a+b\right)^3\)

Có: \(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge2\sqrt{\frac{\left(a+b\right)^6}{4}}=\left(a+b\right)^3\)\(\Rightarrow\)\(VT^2\ge\frac{\left(a+b\right)^3}{2}\) (2) 

(1) và (2) => \(VT^2\ge VP^2\) => \(VT\ge VP\) ( đpcm ) 

ta có:(a-b)²(17a²+10ab+9b²)>(=)0

\(\Leftrightarrow\sqrt{2a\left(a+b\right)^3}\le\frac{5}{2}a^2+\frac{3}{2}b^2\)

áp dụng cô si ta có:

\(2b\sqrt{2\left(a^2+b^2\right)}\le\frac{4b^2+2\left(a^2+b^2\right)}{2}\)

\(\Rightarrow b\sqrt{2\left(a^2+b^2\right)}\le\frac{1}{2}a^2+\frac{3}{2}b^2\)

\(\Rightarrow\sqrt{2a\left(a+b\right)^3}+b\sqrt{2\left(a^2+b^2\right)}\le\frac{5}{2}a^2+\frac{1}{2}a^2+\frac{3}{2}b^2+\frac{3}{2}b^2=3\left(a^2+b^2\right)\)

cho x+y+z=4 

cmr \(\frac{1}{xy}+\frac{1}{yz}\ge1\)

BL

TA CẦN CM \(\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge1\Leftrightarrow\frac{1}{y}+\frac{1}{z}\ge x\)

mà x=\(4-\left(y+z\right)\)

\(\Rightarrow\frac{1}{y}+\frac{1}{z}\ge4-\left(y+z\right)\Leftrightarrow\frac{1}{y}-2+y+\frac{1}{z}-2+z\ge0\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{y}}-\sqrt{y}\right)^2+\left(\frac{1}{\sqrt{z}}-\sqrt{z}\right)^2\ge0\)(luôn đúng)

+Tuấn 10B_2 (T ko biết đánh word nên dùng tạm .V)

GPT: \(\(\sqrt{x+3}+\sqrt[3]{x}=3\)\) (Bài này cách lp 9 dễ t ko giải nữa)

Vì \(\(f\left(x\right)=\sqrt{x+3}+\sqrt[3]{x}=3\)\) là hàm tăng trên tập [-3;\(\(+\infty\)\))

Ta có: Nếu \(\(x>1\Leftrightarrow f\left(x\right)>f\left(1\right)=3\)\)nên pt vô nghiệm

Nếu \(\(-3\le x< 1\Leftrightarrow f\left(x\right)< f\left(1\right)=3\)\)nên pt vô nghuêmj

Vậy x = 1

B2, GHPT: \(\(\hept{\begin{cases}2x^2+3=\left(4x^2-2yx^2\right)\sqrt{3-2y}+\frac{4x^2+1}{x}\\\sqrt{2-\sqrt{3-2y}}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\end{cases}}\)\)

ĐK \(\(\hept{\begin{cases}-\frac{1}{2}\le y\le\frac{3}{2}\\x\ne0\\x\ne-\frac{1}{2}\end{cases}}\)\)

Xét pt (1) \(\(\Leftrightarrow2x^2+3-4x-\frac{1}{x}=x^2\left(4-2y\right)\sqrt{3-2y}\)\)

\(\(\Leftrightarrow-\frac{1}{x^3}+\frac{3}{x^2}-\frac{4}{x}+2=\left(4-2y\right)\sqrt{3-2y}\)\)

\(\(\Leftrightarrow\left(-\frac{1}{x}+1\right)^3+\left(-\frac{1}{x}+1\right)=\left(\sqrt{3-2y}\right)^3+\sqrt{3-2y}\)\)

Xét hàm số \(\(f\left(t\right)=t^3+t\)\)trên R có \(\(f'\left(t\right)=3t^2+1>0\forall t\in R\)\)

Suy ra f(t) đồng biến trên R . Nên \(\(f\left(-\frac{1}{x}+1\right)=f\left(\sqrt{3-2y}\right)\Leftrightarrow-\frac{1}{x}+1=\sqrt{3-2y}\)\)

Thay vào (2) \(\(\sqrt{2-\left(1-\frac{1}{x}\right)}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\)\)

\(\(\Leftrightarrow\sqrt{\frac{1}{x}+1}=\frac{\sqrt[3]{x^2\left(x+2\right)}+x+2}{2x+1}\)\)

\(\(\Leftrightarrow\left(2x+1\right)\sqrt{\frac{1}{x}+1}=x+2+\sqrt[3]{x^2\left(x+2\right)}\)\)

\(\(\Leftrightarrow\left(2+\frac{1}{x}\right)\sqrt{1+\frac{1}{x}}=1+\frac{2}{x}+\sqrt[3]{1+\frac{2}{x}}\)\)

\(\(\Leftrightarrow f\left(\sqrt{1+\frac{1}{x}}\right)=f\left(\sqrt[3]{1+\frac{2}{x}}\right)\)\)

\(\(\Leftrightarrow\sqrt{1+\frac{1}{x}}=\sqrt[3]{1+\frac{2}{x}}\)\)

\(\(\Leftrightarrow\left(1+\frac{1}{x}\right)^3=\left(1+\frac{2}{x}\right)^2\)\)

Đặt \(\(\frac{1}{x}=a\)\)

\(\(\Rightarrow Pt:\left(a+1\right)^3=\left(2a+1\right)^2\)\)

Tự làm nốt , mai ra lớp t giảng lại cho ...

Bài 1:

a)

ta có: \(\dfrac{50}{100}=\dfrac{1}{2};\dfrac{-\dfrac{4}{13}}{-\dfrac{8}{13}}=\dfrac{1}{2};\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{1}{2};\dfrac{-\dfrac{2}{17}}{-\dfrac{4}{17}}=\dfrac{1}{2}\)

\(\dfrac{50}{100}=\dfrac{\dfrac{4}{13}}{\dfrac{8}{13}}=\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{\dfrac{2}{17}}{\dfrac{4}{17}}=\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}=\dfrac{1}{2}\)

vậy \(A=\dfrac{1}{2}\)

b)

\(B=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\\ B=\dfrac{1}{19}-\dfrac{1}{19}+\dfrac{2}{29}-\dfrac{2}{29}+\dfrac{3}{39}-...-\dfrac{199}{1999}+\dfrac{200}{2009}\\ B=\dfrac{200}{2009}\)

Bài 2:

\(\dfrac{a}{b}=\dfrac{b}{3c}=\dfrac{c}{9a}=\dfrac{b+c}{3c+9a}\)

suy ra: \(b=\dfrac{3c\left(b+c\right)}{3c+9a}=\dfrac{3cb+3c^2}{3c+9a}=\dfrac{bc+c^2}{c+3a}\)

\(c=\dfrac{9a\left(b+c\right)}{3c+9a}=\dfrac{9ab+9ac}{3c+9a}=\dfrac{3ab+3ac}{c+3a}\)

giả sử b=c là đúng thì :\(\dfrac{bc+c^2}{c+3a}=\dfrac{3ab+3ac}{c+3a}\)

hay \(bc+c^2=3ab+3ac\\ \Leftrightarrow c^2+bc-3ab-3ac=0\)

\(\Leftrightarrow\left(b+c\right)\left(c-3a\right)=0\Rightarrow c-3a=0\Rightarrow c=3a\)

b) \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}+\dfrac{2}{2014.2016}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2016}\right)=\dfrac{2015}{4032}< 1\)

mà \(1< \dfrac{4}{3}\) nên \(\dfrac{2015}{4032}< \dfrac{4}{3}\)

hay \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}< \dfrac{4}{3}\)

bài 3:

a)\(\left(x-y\right)\left(x+y\right)=x^2-y^2-xy+xy=x^2-y^2\) (đpcm)

b) áp dụng BĐT tam giác, ta có:

\(a+b>c\Rightarrow a+b-c>0\\ b+c>a\Rightarrow b+c-a< 0\\ a+c>b\Rightarrow a-b+c>0\)

suy ra: \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< 0­\: ­\: ­\: ­\: ­\: ­\: \)

đồng thời \(abc>0\) với mọi a, b, c dương.

nên \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< abc\)

ko tìm dc dấu bằng xảy ra.

Đây là đề bài:

Kiểm tra hộ mik lời giải, nếu có cách khác các bn góp ý cho mik nha, thnks nhiều!

Có \(P=\dfrac{2}{x^2+y^2}+\dfrac{35}{xy}+2xy\\ \Leftrightarrow P=\left(\dfrac{2}{x^2+y^2}+\dfrac{1}{xy}\right)+\dfrac{2}{xy}+\left(\dfrac{32}{xy}+2xy\right)\)

Xét nhóm 1: Áp dụng BĐT\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\left(1\right)\ge2\left(\dfrac{4}{\left(x+y\right)^2}\right)\ge2\left(\dfrac{4}{4^2}\right)=\dfrac{1}{2}\Rightarrow Min\left(1\right)=\dfrac{1}{2}\Leftrightarrow x=y\\\)

Xét nhóm 2: Vì \(x+y\le4\Rightarrow2\sqrt{xy}\le4\Rightarrow xy\le4\Rightarrow\dfrac{1}{xy}\ge\dfrac{1}{4}\Rightarrow Min\left(2\right)=\dfrac{1}{2}\Leftrightarrow xy=4\\ \)

Xét nhóm 3:Áp dụng BĐT Cô-si ta được:\(\dfrac{32}{xy}+2xy\ge2\sqrt{\dfrac{32}{xy}\cdot2xy}=16\Rightarrow Min\left(3\right)=16\Leftrightarrow x=y\\ \)

Từ các NX trên\(\Rightarrow MinP=\dfrac{1}{2}+\dfrac{1}{2}+16=17\left(ĐK:\right)x=y;xy=4hayx=y=2\)