Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=-15,5\left(20,8+9,2\right)+3,5\left(9,2+20,8\right)\)
\(=30\cdot\left(-12\right)=-360\)
b: \(=\dfrac{21}{20}:\dfrac{27}{10}+2+\left(\dfrac{2}{5}:\dfrac{5}{2}\right)\cdot\left(\dfrac{21}{5}-\dfrac{13}{10}\right)\)
\(=\dfrac{21}{10}\cdot\dfrac{10}{27}+2+\dfrac{4}{25}\cdot\dfrac{29}{10}\)
\(=\dfrac{7}{9}+2+\dfrac{58}{125}=\dfrac{3647}{1125}\)
A) -15,5 x 20,8 +2,5 x 9,2 -15,5 x 9,2 + 3,5 x 20,8
=-15,5 x 20,8 + 2,5 x 9,2 + -15,5 x 9,2 +3,5 x 20,8
=[20,8x(-15,5+3,5)]+[9,2x(2,5+-15,5)]
=[20,8x(-12)]+[9,2x(-13)]
=-249,6+-119,6
=-369,2
Bài 3:
a: \(S=1+5^2+5^4+\cdots+5^{200}\)
=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)
=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)
=>24S=\(5^{202}-1\)
=>\(S=\frac{5^{202}-1}{24}\)
b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)
\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)
mà \(4^{15}>3^{11}\)
nên \(4^{30}>3\cdot24^{10}\)
=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)
Bài 2:
a: |2x-3|>5
=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)
c: |3x-1|<=7
=>-7<=3x-1<=7
=>-6<=3x<=8
=>\(-2\le x\le\frac83\)
d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)
TH1: \(x<-\frac32\)
=>2x+3<0; 3x-5<0
(1) sẽ trở thành: -2x-3-3x+5=7
=>-5x+2=7
=>-5x=5
=>x=-1(loại)
TH2: -3/2<=x<5/3
=>2x+3>=0; 3x-5<0
(1) sẽ trở thành: 2x+3-3x+5=7
=>-x+8=7
=>-x=-1
=>x=-1(nhận)
TH3: x>=5/3
=>2x+3>0; 3x-5>=0
(1) sẽ trở thành: 2x+3+3x-5=7
=>5x-2=7
=>5x=9
=>x=9/5(nhận)
a) Ta có : (1/16)10 = [(1/2)4]10 = (1/2)40
Vì (1/2)40 < (1/2)50 nên (1/16)10 < (1/2)50
b) Ta có : 430 = ( 2 . 2)30 = 230 . 230 = (22)15 . (23)10 > 315 . 810 > 3 . 310 .810 = 3 . (3 . 8)10 = 3 .2410
Vậy nên 230 + 330 + 430 > 2410 . 3
Mình chỉ giải thế thôi, còn đâu bn tự làm tiếp
2)
a) \(2\left|2x-3\right|=1\)
=> \(\left|2x-3\right|=1:2\)
=> \(\left|2x-3\right|=\frac{1}{2}\)
=> \(\left[{}\begin{matrix}2x-3=\frac{1}{2}\\2x-3=-\frac{1}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}2x=\frac{1}{2}+3=\frac{7}{2}\\2x=\left(-\frac{1}{2}\right)+3=\frac{5}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\frac{7}{2}:2\\x=\frac{5}{2}:2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{5}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{7}{4};\frac{5}{4}\right\}.\)
b) \(7,5-3\left|5-2x\right|=-4,5\)
=> \(4,5\left|5x-2\right|=-4,5\)
=> \(\left|5x-2\right|=\left(-4,5\right):4,5\)
=> \(\left|5x-2\right|=-1\)
Ta luôn có: \(\left|x\right|>0\forall x\)
=> \(\left|5x-2\right|>-1\)
=> \(\left|5x-2\right|\ne-1\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
c) \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có: \(\left|3x-4\right|>\) hoặc \(=0\forall x\)
\(\left|3y+5\right|>\) hoặc \(=0\forall y.\)
=> \(\left|3x-4\right|+\left|3y+5\right|=0\)
=> \(\left[{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}3x=0+4=4\\3y=0-5=-5\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=4:3\\y=\left(-5\right):3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{4}{3}\\y=-\frac{5}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{4}{3}\right\};y\in\left\{-\frac{5}{3}\right\}.\)
Chúc bạn học tốt!
Bài 1:
a) \(-15,5.20,8+3,5.9,2-15,5.9,2+3,5.20,8\)
\(=20,8.\left(-15,5+3,5\right)+9,2.\left(-15,5+3,5\right)\)
\(=\left(-15,5+3,5\right).\left(20,8+9,2\right)\)
\(=\left(-12\right).30=-360\)
b) \(\left[\left(-19,95\right)+\left(-45,75\right)\right]+\left[4,95+5,75\right]\)
\(=\left[\left(-19,95\right)+4,95\right]+\left[\left(-45,75\right)+5,75\right]\)
\(=-15+\left(-40\right)=-55\)
Bài 2 :
\(a,2.\left|2x-3\right|=1\)
\(\Leftrightarrow\left|2x-3\right|=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\frac{1}{2}\\2x-3=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{1}{2}+3\\2x=-\frac{1}{2}+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{7}{2}\\2x=\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{5}{4}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{7}{4},\frac{5}{4}\right\}\)
\(b,7.5-3\left|5-2x\right|=-4.5\)
\(\Leftrightarrow3.\left|5-2x\right|=7.5-\left(-4.5\right)=12\)
\(\Leftrightarrow\left|5-2x\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{2},\frac{9}{2}\right\}\)
\(c,\left|3x-4\right|+\left|3y+5\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=-\frac{5}{3}\end{matrix}\right.\)
Vậy : \(\left(x,y\right)=\left(\frac{4}{3},-\frac{5}{3}\right)\)
Bài 3 :
a) \(2^{300}\) và \(3^{200}\)
Ta có : \(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
mà : \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
Vậy : \(3^{200}>2^{300}\)
b) \(2^{30}+3^{30}+4^{30}\) và \(3.2.4^{10}\)
Ta có : \(3.2.4^{10}=6.\left(2^2\right)^{10}=6.2^{20}=3.2^{21}\)
Ta thấy : \(2^{30}>3.2^{21}\Rightarrow2^{30}+3^{30}+4^{30}>3.2^{21}\)
hay : \(2^{30}+3^{30}+4^{30}>3.2.4^{10}\)
Vậy : \(2^{30}+3^{30}+4^{30}>3.2.4^{10}\)
Chúc bạn học tốt !