\(•nếu\:x\ge0\:thì\:\left|x.\dfrac{3}{5}\right|=\dfrac{3x}{5}\\ •nếu\: x< 0\: thì\: \left|x.\dfrac{3}{5}\right|=-\dfrac{3x}{5}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{5}< \dfrac{1}{3}\left(với\:x\ge0\right)\\-\dfrac{3x}{5}< \dfrac{1}{3}\left(với\:x< 0\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x< \dfrac{5}{9}\left(nhân\:0\le x< \dfrac{5}{9}\right)\\x>-\dfrac{5}{9}\left(nhận\:-\dfrac{5}{9}>x\ge0\right)\end{matrix}\right.\)

vậy pht có nghiệm là \(-\dfrac{5}{9}< x< \dfrac{5}{9}\)

\(a\text{)}\:x^2-2\left(\dfrac{1}{2}+2\right)x+\dfrac{1}{2}+1=0\\ x^2-5x+\dfrac{5}{2}=0\\ x^2-2.\dfrac{5}{2}x+\left(\dfrac{5}{2}\right)^2=\left(\dfrac{5}{2}\right)^2-\dfrac{5}{2}\\ \left(x-\dfrac{5}{2}\right)^2=\dfrac{15}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{15}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{15}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{15}+5}{2}\\x=\dfrac{5-\sqrt{15}}{2}\end{matrix}\right.\)

1)\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\\ A=\left|x-1\right|+\left|x+1\right|\\ A=\left|1-x\right|+\left|x+1\right|\ge\left|1-x+x+1\right|=2\)

dấu "=" xảy ra khi \(\left[{}\begin{matrix}\left\{{}\begin{matrix}1-x\ge0\\x+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-x< 0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1\ge x\\x\ge-1\end{matrix}\right.\left(nhận\right)\\\left\{{}\begin{matrix}1< x\\x< -1\end{matrix}\right.\left(loại\right)\end{matrix}\right.\)

vậy....

\(B=\sqrt{4x^2-12x+9}+\sqrt{4x^2+12x+9}\\ B=\left|2x-3\right|+\left|2x+3\right|\\ B=\left|3-2x\right|+\left|2x+3\right|\ge\left|3-2x+2x+3\right|=6\)

dấu " = " xảy ra khi \(\left[{}\begin{matrix}\left\{{}\begin{matrix}3-2x\ge0\\2x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3-2x< 0\\2x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3\ge2x\\2x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}3< 2x\\2x< -3\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\dfrac{3}{2}\ge x\\x\ge-\dfrac{3}{2}\end{matrix}\right.\left(nhận\right)\\\left\{{}\begin{matrix}\dfrac{3}{2}< x\\x< -\dfrac{3}{2}\end{matrix}\right.\left(loại\right)\end{matrix}\right.\)

vậy....

2)

\(A=\sqrt{x+4}+\sqrt{4-x}\\ A^2=x+4+4-x+2\sqrt{\left(x+4\right)\left(4-x\right)}\\ A^2=4+2\sqrt{16-x^2}\\ vìx^2\ge0nên\\ A^2\le12\\ A\le\sqrt{12}\)

dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x^2\ge0\\x^2\le16\end{matrix}\right.\Rightarrow0\le x\le4\)

vậy...

\(B=\sqrt{x+6}+\sqrt{6-x}\\ B^2=x+6+6-x+2\sqrt{\left(x+6\right)\left(6-x\right)}\\ B^2=12+2\sqrt{36-x^2}\\ vì\: x^2\ge0nên\\ B^2\le24\\ B\le\sqrt{24}\)

dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x^2\ge0\\x^2\le36\end{matrix}\right.\Rightarrow0\le x\le6\)

a+b+c = 10+102+103 = 215

vậy a+b+c = 215

Nửa chu vi là: 104:2 = 52 (cm)

Chiều dài hơn chiều rộng là: 2 x 2 = 4 (cm)

Chiều dài là: (52 + 4):2 = 28

\(1\text{)} \left(x-3\right)-9.\left(2x-7\right)=16:2\\ x-3-18x+63=8\\ -17x=-52\\ x=\dfrac{52}{17}\)

\(2\text{)} x-\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\\ -\dfrac{1}{6}x=\dfrac{5}{12}\\ x=-2,5\)

\(3\text{)} -2.\left(x-1\right)+5.\left(x-2\right)=x-14\\ -2x+2+5x-10=x-14\\ 2x=-6\\ x=-3\)

\(4\text{)} x.\left(x+1\right)-x-1=0\\ \left(x+1\right)\left(x-1\right)=0\\ x^2-1=0\\ x^2=1\\ x=\pm1\)

\(a\text{)} A=x^2-y^2=\left(x-y\right)\left(x+y\right)\\ A=\left(87-13\right)\left(87+13\right)=74.100=7400\)

\(b\text{)} B=9x^2-6x+1=\left(3x-1\right)^2\\ B=\left(3.\dfrac{1}{3}-1\right)^2=0\)

\(c\text{)} C=4x^2-12xy+9y^2=\left(2x-3y\right)^2\\ C=\left(2.1-3.2\right)^2=16\)

\(a\text{)} x^2-11=0\\ x^2=11\\ x=\pm\sqrt{11}\)

\(b\text{)}\:x^2-2\sqrt{13x}+13=0\\ \left(x-\sqrt{13}\right)^2=0\\ x-\sqrt{13}=0\\ x=\sqrt{13}\)

\(c\text{)}\:\sqrt{x^2-10x+25}=7-2x\\ \left|x-5\right|=7-2x\\ \Rightarrow\left[{}\begin{matrix}x-5=7-2x\left(với\:x\ge5\right)\\5-x=7-2x\left(với\:x< 5\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

mình thích hentai

gọi x(m) là độ dài của chiều rộng (x là số tự nhiên)

khi đó chiều dài là : 1,5x (m0

theo đề bài, ta có phương trình:

\(\left(x-5\right)\left(1,5x-5\right)=1,5x^2-1,5x^2.16\%\\ 1,5x^2-5x-7,5x+25=1,5x^2-0,24x^2\\ 0,24x^2-12,5x+25=0\\ 24x^2-1250x+2500=0\\ 24x^2-1200x-50x+2500=0\\ 24x\left(x-50\right)-50\left(x-50\right)=0\\ \left(x-50\right)\left(24x-50\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-50=0\\24x-50=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=50\left(nhận\right)\\x=\dfrac{50}{24}\left(loại\right)\end{matrix}\right.\)

vậy chiều dài là 75m
chiều rộng là 50m