Sửa đề : a) Tìm GTNN A

a) \(A=\left|x-5\right|+3\)có : \(\left|x-5\right|\ge0\Rightarrow\left|x-5\right|+3\ge0\)

\(\Leftrightarrow A\ge3\)dấu "=" xảy ra khi : \(\left|x-5\right|=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)

Vậy GTNN A = 3 khi x = 5.

b) \(C=-\left|x+1\right|+5\)có : \(-\left|x+1\right|\le0\Rightarrow-\left|x+1\right|+5\le5\)

\(\Leftrightarrow C\le5\)dấu "=" xảy ra khi : \(-\left|x+1\right|=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy GTLN C = 5 khi x = -1.

\(D=5-\left|2x+3\right|\)có : \(-\left|2x+3\right|\le0\Rightarrow5-\left|2x+3\right|\le5\)

\(\Leftrightarrow D\le5\)dấu "=" xảy ra khi : \(-\left|2x+3\right|=0\Leftrightarrow2x+3=0\Leftrightarrow x=-\frac{3}{2}\)

Vậy GTLN D = 5 khi x = -3/2.

c) \(\left|x-3\right|+\left|y+1\right|=0\)có \(\left|x-3\right|\ge0;\left|y+1\right|\ge0\Rightarrow\left|x-3\right|+\left|y+1\right|\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=0\\\left|y+1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}.\)

• Đỗ Đức Lợi ơi

• B=|2x+1|-4 

a)\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

   \(\frac{1}{3}:2x=-5-\frac{1}{4}\)

   \(\frac{1}{3}:2x=-\frac{21}{3}\)

   \(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)

   \(2x=-\frac{1}{21}\)

   \(x=\frac{-1}{42}\)

b)\(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)

c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

   \(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)

a) 1/4 + 1/3 : 2x = -5

=> 1/3 : 2x = -5 - 1/4

=> 1/3 : 2x = -21/4

=> 2x = 1/3 : (-21/4) = -4/63

=> x = -4/63 : 2 = -2/63

2x-1=4

2x-5=4

2x=4+1

2x=4+5

2x=5

2x=9

x=5/2

x=9/2

x=2.5

x=4.5

\(\left|2x-1\right|+\left|2x-5\right|=4\)

\(\Leftrightarrow\left|2x-1\right|+\left|5-2x\right|=4\)

Ta có: \(\hept{\begin{cases}\left|2x-1\right|\ge2x-1\forall x\\\left|5-2x\right|\ge5-2x\forall x\end{cases}}\)

\(\Rightarrow\left|2x-1\right|+\left|2x-5\right|\ge\left(2x-1\right)+\left(5-2x\right)=2x-1+5-2x=4\)

Mà \(\left|2x-1\right|+\left|2x-5\right|=4\)

\(\Rightarrow\hept{\begin{cases}\left|2x-1\right|=2x-1\\\left|5-2x\right|=5-2x\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1\ge0\\5-2x\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge\frac{1}{2}\\x\le\frac{5}{2}\end{cases}\Rightarrow}\frac{1}{2}\le x\le\frac{5}{2}}\)

Vậy \(\frac{1}{2}\le x\le\frac{5}{2}\)

Tham khảo nhé~

a, 11/12 - ( 2/5 + x ) = 2/3

<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)

=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)

b, 2x . ( x - 1/7 ) = 0

<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

vậy x={\(0;\frac{1}{7}\)}

c, 3/4 + 1/4 : x = 2/5

<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)

vậy x=-5/7

a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)

\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)

\(\Leftrightarrow x=-\frac{3}{20}\)

b) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

\(\Leftrightarrow4x=\frac{-20}{7}\)

\(\Leftrightarrow x=-\frac{5}{7}\)

=> |2x+3| = 5+2.|4-x| = 5+|8-2x|

=> 2x+3 = 5+8-2x hoặc 2x+3 = 5-8+2x

=> x = 5/2 

Vậy x = 5/2

Tk mk nha

1) \(\left|2x+5\right|\ge21\Rightarrow2x+5\ge21\)hoặc \(2x+5

2b) Áp dụng bất đẳng thức giá trị tuyệt đối: |a| + |b|  \(\ge\) |a + b|. Dấu "=" xảy ra khi tích a.b \(\ge\) 0 

Ta có: B = |2x - 1| + |3 - 2x| + 5  \(\ge\) |2x - 1+3 - 2x| + 5  = |2| + 5 = 7

=> Min B = 7 khi

(2x - 1)( 3 - 2x) \(\ge\) 0 => (2x - 1)(2x - 3) \(\le\) 0 

Mà 2x - 1 > 2x - 3 nên 2x - 1 \(\ge\) 0 và 2x - 3 \(\le\)  0 

=> x \(\ge\) 1/2 và x  \(\le\) 3/2