a) \(\Rightarrow\left(x-3\right)\left(x+4\right)=5.12\)

\(\Rightarrow x^2+x-72=0\)

\(\Rightarrow\left(x-8\right)\left(x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)

b) \(\Rightarrow\left(x+3\right)^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x+3=6\\x+3=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-9\end{matrix}\right.\)

c) \(\Rightarrow2x^2=8\Rightarrow x^2=4\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

Làm hết á!

b6, 

aTa có: |2x+4|≥0 với mọi x

⇒|2x+4|-5≥-5

Dấu "=" xảy ra ⇌ |2x+4|=0⇒x=-2

vậy GTNN P= -5 ⇌ x=-2

b, ta có (2x+7)\(^2\)≥0với mọi x

⇒(2x+7)\(^2\)+\(\dfrac{2}{5}\)≥\(\dfrac{2}{5}\)

Dấu "=" xảy ra⇌ x=-\(\dfrac{7}{2}\)

vậy GTNN E= \(\dfrac{2}{5}\)⇌x=-\(\dfrac{7}{2}\)

b7, ta có x\(^2\)≥0 với mọi x

⇒-x\(^2\)≤0 với mọi x

⇒-x\(^2\)-5≤-5

Dấu "=" xảy ra ⇌x=0

vậy GTLN A=-5⇌x=0

\(1,\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-3\right)^2-4\left(-2\right)\left(-m+1\right)>0\\x_1+x_2=\dfrac{3}{-2}< 0\\x_1x_2=\dfrac{-m+1}{-2}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}17-8m>0\\-m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{17}{8}\\m>1\end{matrix}\right.\Leftrightarrow1< m< \dfrac{17}{8}\)

\(2,\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-4\right)^2-4\left(-3\right)\left(-2m+1\right)\ge0\\x_1+x_2=\dfrac{4}{-3}< 0\\x_1x_2=\dfrac{-2m+1}{-3}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28-24m\ge0\\-2m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{7}{6}\\m>\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< m\le\dfrac{7}{6}\)

4.

\(sinx+2cos\left(x+\dfrac{\pi}{3}\right)+4sin\left(x+\dfrac{\pi}{6}\right)+cosx=4\)

\(\Leftrightarrow sinx+cosx-\sqrt{3}sinx+2\sqrt{3}sinx+2cosx+cosx=4\)

\(\Leftrightarrow\left(1+\sqrt{3}\right)sinx+4cosx=4\)

\(\Leftrightarrow\sqrt{20+2\sqrt{3}}\left(\dfrac{1+\sqrt{3}}{\sqrt{20+2\sqrt{3}}}sinx+\dfrac{4}{\sqrt{20+2\sqrt{3}}}cosx\right)=4\)

\(\Leftrightarrow cos\left(x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}\right)=\dfrac{4}{\sqrt{20+2\sqrt{3}}}\)

\(\Leftrightarrow x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}=\pm arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)

3.

\(4sinx+cosx+2cos\left(x+\dfrac{\pi}{3}\right)=2\)

\(\Leftrightarrow4sinx+cosx+cosx-\sqrt{3}sinx=2\)

\(\Leftrightarrow\left(4-\sqrt{3}\right)sinx+2cosx=2\)

\(\Leftrightarrow\sqrt{23-4\sqrt{3}}\left(\dfrac{4-\sqrt{3}}{\sqrt{23-4\sqrt{3}}}sinx+\dfrac{2}{\sqrt{23-4\sqrt{3}}}cosx\right)=2\)

\(\Leftrightarrow cos\left(x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}\right)=\dfrac{2}{\sqrt{23-4\sqrt{3}}}\)

\(\Leftrightarrow x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}=\pm arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)

a: Xét ΔHAC vuông tại H và ΔABC vuông tại A có

\(\widehat{C}\) chung

Do đó: ΔHAC~ΔABC

b: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=15^2+20^2=625\)

=>BC=25

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}BH\cdot BC=BA^2\\AH\cdot BC=AB\cdot AC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH\cdot25=15^2=225\\AH\cdot25=15\cdot20=300\end{matrix}\right.\)

=>BH=9; AH=12

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10