Toán

\(a,2\sqrt{50}-\sqrt{108}-\sqrt{98}+2\sqrt{27}\)

\(=2\sqrt{25\cdot2}-\sqrt{36\cdot3}-\sqrt{49\cdot2}+2\sqrt{9\cdot3}\)

\(=2\cdot5\sqrt{2}-6\sqrt{3}-7\sqrt{2}+2\cdot3\sqrt{3}\)

\(=10\sqrt{2}-6\sqrt{3}-7\sqrt{2}+6\sqrt{3}\)

\(=3\sqrt{2}\)

\(72-3x=6^2\)

\(\Rightarrow72-3x=36\)

\(\Rightarrow3x=72-36\)

\(\Rightarrow3x=36\)

\(\Rightarrow x=\dfrac{36}{3}=12\)

\(Vậy:x=12\)

\(\Leftrightarrow2x\left(1,5x+1\right)-3\left(2-x\right)^2\ge6.5x-2.12\)

\(\Leftrightarrow3x^2+2x-12+12x-3x^2\ge30x-24\)

\(\Leftrightarrow16x\le12\)

\(\Leftrightarrow x\le\dfrac{3}{4}\)

\(\dfrac{x\left(1,5x+1\right)}{6}-\dfrac{\left(2-x\right)^2}{4}\ge\dfrac{5x}{2}-2\)

\(\Leftrightarrow\dfrac{2x\left(1,5x+1\right)}{12}-\dfrac{3\left(4-4x+x^2\right)}{12}\ge\dfrac{6\cdot5x-2\cdot12}{12}\)

\(\Leftrightarrow3x^2+2x-12+12x-3x^2\ge30x-24\)

\(\Leftrightarrow14x-12\ge30x-24\)

\(\Leftrightarrow30x-14x\ge-12+24\)

\(\Leftrightarrow16x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{16}\)

\(\Leftrightarrow x\le\dfrac{3}{4}\)

Vậy: ..

Bài 14:

a) \(x^2+6x+9\)

\(=x^2+2\cdot3\cdot x+3^2\)

\(=\left(x+3\right)^2\)

b) \(x^2-10x+25\)

\(=x^2-2\cdot5\cdot x+5^2\)

\(=\left(x-5\right)^2\)

c) \(4x^2+4x+1\)

\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)

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

d) \(9x^2-12x+4\)

\(=\left(3x\right)^{^2}-2\cdot3x\cdot2+2^2\)

\(=\left(3x-2\right)^2\)

e) \(x^2+2xy+y^2\)

\(=\left(x+y\right)^2\)

f) \(9x^2+24x+16\)

\(=\left(3x\right)^2+2\cdot3x\cdot4+4^2\)

\(=\left(3x+4\right)^2\)

h) \(x^2-x+\dfrac{1}{4}\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2\)

\(=\left(x-\dfrac{1}{2}\right)^2\)

i) \(\left(x+y\right)^2-2\cdot\left(x+y\right)\cdot z+z^2\)

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

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

Hai tọa độ \(y=\left(m+2\right)x+3;y=\left(4m-1\right)x+1\) cắt nhau khi :

\(\left(m+2\right)x+3=\left(4m-1\right)x+1\left(1\right)\)

Thay \(x=2\) vào \(\left(1\right)\), ta được :

\(\left(m+2\right).2+3=\left(4m-1\right).2+1\\ \Leftrightarrow2m+4+3=8m-2+1\)

\(\Leftrightarrow2m-8m=-1-7\\ \Leftrightarrow-6m=-8\\ \Leftrightarrow m=\dfrac{4}{3}\)

Vậy \(m=\dfrac{4}{3}\) thi thỏa mãn đề bài.

Gọi \(A\left(x;y\right)\) là tọa độ giao điểm của 2 đồ thị.

2 đồ thị \(y=2x+7;y=4x-5\) giao nhau khi :

\(2x+7=4x-5\\ \Leftrightarrow2x-4x=-5-7\\ \Leftrightarrow-2x=-12\\ \Leftrightarrow x=6\)

Thay \(x=6\) vào \(y=2x+7\Rightarrow y=2.6+7=19\)

Vậy \(A\left(6;19\right)\)

a) \(9\sqrt{x+2}-\dfrac{1}{3}\sqrt{9x+18}=24\) (ĐK: \(x\ge-2\)) 

\(\Leftrightarrow9\sqrt{x+2}-\dfrac{1}{3}\sqrt{9\left(x+2\right)}=24\)

\(\Leftrightarrow9\sqrt{x+2}-\dfrac{1}{3}\cdot3\sqrt{x+2}=24\)

\(\Leftrightarrow9\sqrt{x+2}-\sqrt{x+2}=24\)

\(\Leftrightarrow8\sqrt{x+2}=24\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{24}{8}\)

\(\Leftrightarrow\sqrt{x+2}=3\)

\(\Leftrightarrow x+2=9\)

\(\Leftrightarrow x=9-2\)

\(\Leftrightarrow x=7\left(tm\right)\)

b) \(\sqrt{x^2-6x+9}-2=0\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\)

\(\Leftrightarrow\left|x-3\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=-2\left(x< 3\right)\\x-3=2\left(x\ge3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3-2\\x=3+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)