Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
\(\left(-3,8\right)+\left[\left(-5,7\right)+\left(+3,8\right)\right]\\ =\left(-3,8\right)+\left(-5,7\right)+3,8\\ =\left[\left(-3,8\right)+3,8\right]+\left(-5,7\right)\\ =0+\left(-5,7\right)\\ =-5,7\)
b)
\(\left(+31,4\right)+\left[\left(+6,4\right)+\left(-18\right)\right]\\ =31,4+6,4-18\\ =37,8-18\\ =19,8\)
c)
\(\left[\left(-9,6\right)+\left(+4,5\right)\right]+\left[\left(+9,6\right)+\left(-1,5\right)\right]\\ =\left(-9,6\right)+4,5+9,6-1,5\\ =\left[\left(-9,6\right)+9,6\right]+\left[4,5-1,5\right]\\ =0+3\\ =3\)
d)
\(\left[\left(-4,9\right)+\left(-37,8\right)\right]+\left[\left(+1,9\right)+\left(+2,8\right)\right]\\ =\left(-4,9\right)-37,8+1,9+2,8\\ =\left[\left(-4,9+1,9\right)\right]-\left[\left(37,8-2,8\right)\right]\\ =\left(-3\right)-35\\ =-38\)
a)(-3,8)+[(-5,7)+3,8]
=(-3,8)+(-5,7)+3,8
=(-3,8)+3,8+(-5,7)
=0+(-5,7)
=-5,7
a)
\(\begin{array}{l}\left( { - \frac{5}{6}} \right) - \left( { - 1,8} \right) + \left( { - \frac{1}{6}} \right) - 0,8\\ = \left( { - \frac{5}{6}} \right) + 1,8 + \left( { - \frac{1}{6}} \right) - 0,8\\ = \left[ {\left( { - \frac{5}{6}} \right) + \left( { - \frac{1}{6}} \right)} \right] + \left[ {1,8 - 0,8} \right]\\ =\frac{-6}{6}+1= - 1 + 1 = 0\end{array}\)
b)
\(\begin{array}{l}\left( { - \frac{9}{7}} \right) + \left( { - 1,23} \right) - \left( { - \frac{2}{7}} \right) - 0,77\\ = \left[ {\left( { - \frac{9}{7}} \right) - \left( { - \frac{2}{7}} \right)} \right] + \left[ {\left( { - 1,23} \right) - 0,77} \right]\\ =\frac{-7}{7}+(-2)= - 1 + \left( { - 2} \right) = - 3\end{array}\)
a) \(...=0,25+1500+\left(0,5\right)^2=0,25+0,25=1500=1500,5\)
b) \(...=2,7-4,4+5,6-7,3=2,7+5,6-4,4-7,3=8,3-11,7=-3,4\)
c) \(...=-5,44+5+0,44=-5,44+0,44+5=-5+5=0\)
d) \(...=6,72+5,27-0,72-1,27=6,72-0,72+5,27-1,27=6+4=10\)
Tìm GTNN của biểu thức:
a) A = |x+5|+|x+17|
Giải
Ta có : A = |x+5|+|x+17| \(\ge\) |x+5+x+17|
A = |-x-5|+|x+17| \(\ge\) |-x-5+x+17| = | -12 | = 12
Dấu bằng xảy ra khi - 17 \(\le\) x \(\le\) -5
Vậy MinA=12 khi - 17 \(\le\) x \(\le\) -5
b) B = |x+8|+|x+13|+|x+50|
Giải
B = |x+8|+|x+13|+|x+50| \(\ge\) (| x+8|+|-50-x |)+|x+13|
= (| x+8-50-x |)+|x+13|
= |-42| + |x+13|
= 42 + |x+13| \(\ge\) 42
Vậy MinB = 42 khi và chỉ khi:
\(\left\{{}\begin{matrix}x+8\ge0\\x+13=0\\x+50\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge-8\\x=-13\\x\ge-50\end{matrix}\right.\) \(\Rightarrow x=-13\)
c) C = |x+5|+|x+2|+|x−7|+|x−8|
Giải
C = |x+5|+|x+2|+|x−7|+|x−8|
\(\ge\) |x+5| + |x+2| + |7-x| + |8-x|
\(\ge\) |x+5+7-x| + |x+2+8-x|
\(\ge\) |12| + |10|
\(\ge\) 12 + 10 \(\ge\) 22
Vậy MinC = 22 khi và chỉ khi :
-5 \(\le\) x \(\le\) 8 và -2 \(\le\) x \(\le\) 7 \(\Leftrightarrow\) -2 \(\le\) x \(\le\) 7
d) D = |x+3|+|x−2|+|x−5|
Giải
D = |x+3|+|x−2|+|x−5|
\(\ge\) ( |x+3|+|5-x| ) + |x-2| \(\ge\) | x+3+5-x | + | x-2 | \(\ge\) | 8 | + | x-2 | \(\ge\) 8 + | x-2 | \(\ge\) 8 Vậy MinD = 8 khi và chỉ khi: \(\left\{{}\begin{matrix}x+3\ge0\\x-2=0\\5-x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge-3\\x=2\\x\le5\end{matrix}\right.\) \(\Rightarrow x=2\)Tìm GTNN của biểu thức:
a) A = |x+5|+|x+17|
Giải
Ta có : A = |x+5|+|x+17| ≥≥|x+5+x+17|
A = |-x-5|+|x+17| ≥ |-x-5+x+17| = | -12 | = 12
Dấu bằng xảy ra khi - 17 ≤ x ≤ -5
Vậy MinA=12 khi - 17 ≤ x ≤ -5
b) B = |x+8|+|x+13|+|x+50|
Giải
B = |x+8|+|x+13|+|x+50| ≥ (| x+8|+|-50-x |)+|x+13|
= (| x+8-50-x |)+|x+13|
= |-42| + |x+13|
= 42 + |x+13| ≥≥42
Vậy MinB = 42 khi và chỉ khi:
x+8 ≥ 0 ⇒x ≥ −8
x+13 = 0 => x = −13 .Vậy x=-13
x+50 ≥ 0 => x ≥ −50
c) C = |x+5|+|x+2|+|x−7|+|x−8|
Giải
C = |x+5|+|x+2|+|x−7|+|x−8|
=> |x+5| + |x+2| + |7-x| + |8-x|
≥ |x+5+7-x| + |x+2+8-x| = |12| + |10| =12 + 10 = 22
Vậy MinC = 22 khi và chỉ khi :
-5 ≤ x ≤ 8 và -2 ≤ x ≤ 7 ⇔ -2 ≤ x ≤ 7
a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)
a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)
b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)
\(A=\left(-5,85\right)+\left\{\left[\left(+41,3\right)+\left(+5\right)\right]+\left(+0,85\right)\right\}\)
\(A=\left(-5,85\right)+\left\{\left[41,3+5\right]+0,85\right\}\)
\(A=\left(-5,85\right)+\left\{41,3+5+0,85\right\}\)
\(A=\left(-5,85\right)+\left\{41,3+5,85\right\}\)
\(A=\left(-5,85\right)+41,3+5,85\)
\(A=\left(-5,85\right)+5,85+41,3\)
\(A=0+41,3\)
\(A=41,3\)
\(B=\left(-87,5\right)+\left\{\left(+87,5\right)+\left[\left(+3,8\right)+\left(-0,8\right)\right]\right\}\)
\(B=\left(-87,5\right)+87,5+3,8+\left(-0,8\right)\)
\(B=\left[\left(-87,5\right)+87,5\right]+\left[3,8+\left(-0,8\right)\right]\)
\(B=0+3\)
\(B=3\)