Đặt A = (1 - 3x)(x + 2)

= x + 2 - 3x² - 6x

= -3x² - 5x + 2

= \(-3\left(x^2+\frac{5}{3}x-\frac{2}{3}\right)=-3\left(x^2+2.\frac{5}{6}x+\frac{25}{36}-\frac{49}{36}\right)\)

\(=-3\left(x+\frac{5}{6}\right)^2+\frac{49}{12}\le\frac{49}{12}\)

Dấu "=" xảy ra <=> x + 5/6 = 0 <=> x = -5/6

Vậy Max A = 49/12 <=> x = -5/6

\(\left(1-3x\right).\left(x+2\right)\)

\(\rightarrow3.\left(\frac{1}{3}-x\right).\left(x+2\right)\)

\(\rightarrow\left(-3\right).\left(x-\frac{1}{3}\right).\left(x+2\right)\)

\(\rightarrow\left(-3\right).\left(x+\frac{5}{6}-\frac{7}{6}\right).\left(x+\frac{5}{6}+\frac{7}{6}\right)\)

\(\rightarrow\left(-3\right).[\left(x+\frac{5}{6}\right)^2-\frac{49}{36}]\)

\(\rightarrow\left(-3\right).\left(x+\frac{5}{6}\right)^2+\frac{49}{36}\le\frac{49}{36}\)

\(\text{Dấu "="}\)\(\Leftrightarrow x=-\frac{5}{6}\)

x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x - 1)² - y²]

= x(x + y - 1)(x - y - 1)

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

\(=x\left(x-1-y\right)\left(x-1+y\right)\)

\(\frac{2x-1}{x-2}< 3\Leftrightarrow\frac{2x-1}{x-2}-3< 0\)

\(\Leftrightarrow\frac{2x-1-3x+6}{x-2}< 0\Leftrightarrow\frac{-x+5}{x-2}< 0\)

TH1 : \(\hept{\begin{cases}-x+5< 0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}-x< -5\\x>2\end{cases}\Leftrightarrow}\hept{\begin{cases}x>5\\x>2\end{cases}\Leftrightarrow}x>5}\)

TH2 : \(\hept{\begin{cases}-x+5>0\\x-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 5\\x< 2\end{cases}\Leftrightarrow}x< 2}\)

Vậy bft có tập nghiệm S = { x | x > 5 ; x < 2 } 

\(\frac{2x-1}{x-2}< 3\) (ĐK:\(x\ne2\))

\(\Leftrightarrow\frac{2x-1}{x-2}.\left(x-2\right)=3\left(x-2\right)\)

\(\Leftrightarrow2x-1< 3\left(x-2\right)\)

\(\Leftrightarrow2x-3x< -6+1\)

\(\Leftrightarrow-x< -5\)

\(\Leftrightarrow x>5\)(TMĐK)

Vậy x > 5 thì \(\frac{2x-1}{x-2}< 3\)

#Học tốt!!!

\(A=\frac{3x^2+4x}{x^2+1}=\frac{4x^2+4-x^2+4x-4}{x^2+1}=\frac{4\left(x^2+1\right)-\left(x-2\right)^2}{x^2+1}=4-\frac{\left(x-2\right)^2}{x^2+1}\ge4\)

Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2

Vậy Max A = 4 <=> x = 2

4^x - 12.2^x + 32 = 0

<=> (2^x)² - 2.6.2^x + 36 - 4 = 0

<=> (2^x - 6)² - 4 = 0

<=> (2^x - 8)(2^x - 4) = 0

<=> \(\orbr{\begin{cases}2^x=8\\2^x=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

Vậy \(x\in\left\{3;2\right\}\)là nghiệm phương trình 

Xét tam giác \(KBC\)có \(A,H,M\)thẳng hàng và lần lượt thuộc các cạnh \(BK,KC,BC\)nên theo định lí Menelaus ta có: 

\(\frac{HC}{HK}.\frac{AK}{AB}.\frac{MB}{MC}=1\Leftrightarrow\frac{HC}{HK}=\frac{AB}{AK}=\frac{BK}{AK}+1=\frac{BC}{AC}+1\Leftrightarrow\frac{HC}{HK}-\frac{BC}{AC}=1\)

KHÓ QUÁ K BÍT LÀM.SORRY BẠN

không sao đâu bạn ơi! ai biết làm thì giải giúp mình nha!! THANKS