\(\left|x+\dfrac{1}{2}\right|=x+1\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=x+1\\-\left(x+\dfrac{1}{2}\right)=x+1\end{matrix}\right.\\ =>\left[{}\begin{matrix}\dfrac{1}{2}=1\left(\text{VÔ NGHIỆM}\right)\\-x+\dfrac{1}{2}=x+1=>x=-\dfrac{3}{4}\left(TM\right)\end{matrix}\right.\)

Ta có: \(\left|x+\dfrac{1}{2}\right|=x+1\)

=>\(\left\{{}\begin{matrix}x+1>=0\\\left(x+\dfrac{1}{2}\right)^2=\left(x+1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-1\\\left(x+\dfrac{1}{2}-x-1\right)\left(x+\dfrac{1}{2}+x+1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-1\\-\dfrac{1}{2}\left(2x+\dfrac{3}{2}\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\2x+\dfrac{3}{2}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-1\\x=-\dfrac{3}{4}\end{matrix}\right.\)

=>\(x=-\dfrac{3}{4}\)

\(-\dfrac{2}{3}+\left(2^{40}\cdot3^{39}\right):\left(8^{13}\cdot9^{15}\right)\\ =-\dfrac{2}{3}+\left(2^{40}\cdot3^{39}\right):\left[\left(2^3\right)^{13}\cdot\left(3^2\right)^{15}\right]\\ =-\dfrac{2}{3}+\left(2^{40}\cdot3^{39}\right):\left[2^{39}\cdot3^{30}\right]\\ =-\dfrac{2}{3}+2\cdot3^9=-\dfrac{2}{3}+39366=\dfrac{118096}{3}\)

a; 3\(x\) = 2y và \(xy=54\)

\(3x=2y\) ⇒ \(\frac{x}{2}=\frac{y}{3}\) = k ⇒\(x=2k;y=3k\)

Thay \(x=2k;y=3k\) vào biểu thức \(xy\) = 54

Ta có: 2k.2k = 54 ⇒k.k = 54: (2 x 3)

k\(^2\) = 54 : 6

k\(^2\) = 9

\(\left[\begin{array}{l}k=-3\\ k=3\end{array}\right.\)

Nếu k = -3 ta có:

\(\begin{cases}x=2.\left(-3\right)\\ y=3.\left(-3\right)\end{cases}\)

\(\begin{cases}x=-6\\ y=-9\end{cases}\)

Nếu k = 3 ta có: \(\begin{cases}x=2.3\\ y=3.3\end{cases}\)

\(\begin{cases}x=6\\ y=9\end{cases}\)

Vậy (\(x;y\)) = (-6; -9); (6; 9)

\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}=\frac{x+y+z+11+12+13}{13+14+15}=\frac{42}{42}=1\)

\(\Rightarrow x+11=13;y+12=14;z+13=15\)

sau đó r tính ra th

x,y=3/4,1

dy/dx=2/3

TH1: \(x< \dfrac{1}{2}\)

Phương trình sẽ trở thành:

\(1-2x+5-2x=6\)

=>6-4x=6

=>4x=0

=>x=0(nhận)

TH2: \(\dfrac{1}{2}< =x< \dfrac{5}{2}\)

Phương trình sẽ trở thành:

\(2x-1+5-2x=6\)

=>4=6(vô lý)

=>\(x\in\varnothing\)

TH3: \(x>=\dfrac{5}{2}\)

Phương trình sẽ trở thành:

2x-1+2x-5=6

=>4x=12

=>x=3(nhận)

|\(x\) + 7| = 2

\(\left[\begin{array}{l}x+7=-2\\ x+7=2\end{array}\right.\)

\(\left[\begin{array}{l}x=-2-7\\ x=2-7\end{array}\right.\)

\(\left[\begin{array}{l}x=-9\\ x=-5\end{array}\right.\)

Vậy \(x\in\left(-9;-5\right)\)

Nếu \(x=-9\) thay vào biểu thức C = \(-\frac57x^2-\frac34x+1\) ta có:

C = - \(\frac57\left(-9\right)^2\) - \(\frac43\left(-9\right)\) + 1

C = - \(\frac57.81+12\) + 1

C = \(-\frac{405}{7}\) + \(\frac{84}{7}+\frac77\)

C = - \(\frac{321}{7}+\frac77\)

C = \(\) - \(\frac{314}{7}\)

Nếu \(x=-5\) thay vào biểu thức C = \(-\frac57x^2-\frac{4}{3x}+1\) ta có:

C = - \(\frac57.\left(-5\right)^2-\frac43.\left(-5\right)+1\)

C = - \(\frac57.25+\frac{20}{3}+1\)

C = \(-\frac{125}{7}+\frac{20}{3}+1\)

C = \(-\frac{375}{21}+\frac{140}{21}+\frac{21}{21}\)

C = - \(\frac{235}{21}+\frac{21}{21}\)

C = - \(\frac{214}{21}\)

= 11/ 5 (− 3/7 − 5/7 ​ )+ (−8/7) ​ ⋅ 6/11

= −8/7 ( 5/11 + 6/11)

​=-8/7 ​

5/11 x (-3/7) + (-5/7) + (-8/7) x 6/11 = (-15/77) + (-5/7) +(-48/77) = ( -9/11) + (- 5/7) = ( - 118/77)

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