#)Giải :

a)\(\left(x-149\right)\div35=270\)

\(\Leftrightarrow x-149=9450\)

\(\Leftrightarrow x=9599\)

\(a,(x-149):35=270\)

\(\Leftrightarrow x-149=9450\)

\(\Leftrightarrow x=9450+149=9599\)

Vậy x = 9599

\(b,(2009-x):29+32=100\)

\(\Leftrightarrow(2009-x):29=68\)

\(\Leftrightarrow2009-x=1972\)

\(\Leftrightarrow x=2009-1972=37\)

Vậy x = 37

\(c,x+472=918\)

\(\Leftrightarrow x=918-472=446\)

Vậy x = 446

\(=x^7\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^7-x+1\right).\)

TL:

\(x^9+x^8+x^7-x^3+1\)1

\(=x^7\left(x^2+x+1\right)-\left(x^3-1\right)\) 

\(=x^7\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\) 

\(=\left(x^7-x+1\right)\left(x^2+x+1\right)\) 

hc tốt

\(A=\)\(x^2+y^2-4x+y+5.\)

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

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

\(\Rightarrow A_{min}=\frac{3}{4}\Leftrightarrow\)\(\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y+\frac{1}{2}\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-\frac{1}{2}\end{cases}}}\)

\(x^2+y^2-4x+y+5=\left(x-2\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow Min=\frac{3}{4}\)Dấu "=" xr \(\Leftrightarrow\hept{\begin{cases}x-2=0\\y+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-\frac{1}{2}\end{cases}}}\)

\(A=-x^2-5y^2+2xy-4x+20y+13\)

\(=-x^2+2xy-y^2-4y^2-4x+4y+16y+13\)

\(=-\left(x^2-2xy+y^2\right)-\left(4y^2-16y+16\right)-\left(4x-4y\right)+29\)

\(=-\left(x-y\right)^2-4\left(y-2\right)^2-4\left(x-y\right)-4+25\)

\(=-\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]-4\left(y-2\right)^2+25\)

\(=-\left(x-y+2\right)^2-4\left(y-2\right)^2+25\)

\(A_{max}=25\Leftrightarrow\hept{\begin{cases}\left(x-y+2\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y+2=0\\y=2\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)

\(B=-7x^2-y^2+4xy+16x-2y+17.\)

\(=-4x^2+4xy-y^2-3x^2+12x-12+4x-2y+29\)

\(=-\left(2x-y\right)^2-3\left(x-2\right)^2+2\left(2x-y\right)^2-1+30\)

\(=-\left[\left(2x-y\right)^2-2\left(2x-y\right)^2+1\right]-3\left(x-2\right)^2+30\)

\(=-\left(2x-y-1\right)^2-3\left(x-2\right)^2+30\)

\(\Rightarrow B_{max}=30\Leftrightarrow\hept{\begin{cases}\left(2x-y-1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-y-1=0\\x=2\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)

\(A=\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}=\frac{a}{b}+\frac{1}{2}.\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}.\)

Áp dụng BĐT Cauchy cho 6 số dương

\(A\ge6\sqrt[6]{\frac{1}{2^2}.\frac{1}{3^3}.\frac{a}{b}.\sqrt{\frac{b}{c}}.\sqrt{\frac{b}{c}}.\sqrt[3]{\frac{c}{a}}.\sqrt[3]{\frac{c}{a}}.\sqrt[3]{\frac{c}{a}}}=6.\sqrt[6]{\frac{1}{108}}>6.\sqrt[6]{\frac{5^6}{12^6}}=\frac{6.5}{12}=\frac{5}{2}\)

\(231-\left(x-6\right)=1339:13\)

\(231-\left(x-6\right)=103\)

\(x-6=231-103\)

\(x-6=128\)

\(x=134\)

231 - (x - 6) = 1339 : 13

=> 231 - (x - 6) = 103 

=> x - 6 = 231 - 103 

=> x - 6 = 128

=> x =    128 + 6

=> x =  134 

Vậy x = 134  

PTHH: (1) 2 KClO3 -to => 2 KCl + 3 O2
(2) 5 O2 +4 P -to => 2 P2O5
Ta có: nP= 6,2/31= 0,2(mol)
=> nO2(1)= nO2(2)= 5/4 . 0,2= 0,25 (mol)
=> nKClO3 = 2/3. 0,25= 1/6 (mol)
=> mKClO3 = 1/6 . 122,5 ≈ 20,417 (g)

A=−x2−12x+3=−(x2+12x+36)+39=−(x+6)2+39≤39

Vậy GTLN của A là 39 khi x = -6

B=7−4x2+4x=−(4x2−4x+1)+8=−(2x−1)2+8≤8

Vậy GTLN của B là 8 khi x = 

~Hok tốt~

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