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Áp dụng công thức sau:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
Ta có: K = |x - 2,13| + |x+ 2,56|
\(\Rightarrow K=\left|2,13-x\right|+\left|x+2,56\right|\ge\left|2,13-x+x+2,56\right|\)
\(\Rightarrow K=\left|2,13-x\right|+\left|x+2,56\right|\ge\left|4,69\right|=4,69\)
Vậy, minK = 4,69 khi x = 2,13 hoặc x = -2,56
Giải :
-3 = -3
2 = 2
2,13 = 2,13
-√2 = -1,414
\(\frac{3}{7}\)= 0,428
⇒ -3 < -√2 < \(\frac{3}{7}\) < 2 < 2,13
CHÚC BẠN HỌC TỐT ^_^
Giải :
-3 = -3
2 = 2
2,13 = 2,13
-√2 = -1,414
\(-\frac{3}{7}\)= -0,428
⇒ -3 < -√2 < \(-\frac{3}{7}\) < 2 < 2,13
CHÚC BẠN HỌC TỐT ^_^
1
2x . 3=3y .4
=> x=2y=>\(\frac{x}{2}=y\Rightarrow\frac{x}{4}=\frac{y}{2}\)
\(\frac{x}{4}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{5}=\frac{x-2y+3z}{4-4+15}=\frac{1}{15}=\)
x=1/15x4=4/15
y=1/15x2=2/15
z=1/15x6=1/10
\(\Rightarrow x-y-z=\frac{4}{15}-\frac{2}{15}-\frac{1}{10}=\frac{1}{30}\)
\(\left(2x-3\right)^2-2\left(3x+1\right)^2=2x\left(x-2\right)+\left(x-1\right)\left(x+2\right)\)
4\(x^2\)-12x+9-2(9\(x^2\)+6x+1)=2\(x^2\)-4x+\(x^2\)+2x-x-2
4\(x^2\)-12x+9-18\(x^2\)-12x-2=2\(x^2\)-4x+\(x^2\)+2x-x-2
(4\(x^2\)-18\(x^2\)-2\(x^2\)-\(x^2\)) +(-12x-12x+4x-2x+x)+(9-2+2)=0
-17\(x^2\)-21x+9=0
c ) \(\left|3,4-2x\right|:\frac{1}{2}=\frac{3}{4}\)
\(\left|3,4-2x\right|=\frac{3}{8}\)
\(\Rightarrow\left[\begin{array}{nghiempt}3,4-2x=\frac{3}{8}\\3,4-2x=-\frac{3}{8}\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=\frac{121}{80}\\x=\frac{151}{80}\end{array}\right.\)
Vậy ...........
e) \(\left|x+2,8\right|=3,5\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2,8=3,5\\x+2,8=-3,5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0,7\\x=-6,3\end{array}\right.\)
Vậy ....................
f ) \(2.\left|2x-3\right|=\frac{1}{2}\)
\(\left|2x-3\right|=\frac{1}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=\frac{13}{8}\\x=\frac{11}{8}\end{array}\right.\)
Vậy ................
g ) \(7,5-3.\left|5-2x\right|=4,5\)
\(3.\left|5-2x\right|=4,5\)
\(\left|5-2x\right|=\frac{3}{2}\)
\(\Rightarrow\left[\begin{array}{nghiempt}5-2x=\frac{3}{2}\\5-2x=-\frac{3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=\frac{7}{4}\\x=\frac{13}{4}\end{array}\right.\)
Vậy ..............