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a) (x+5)^5=2^10 =>(x+5)^5=4^5 =>x+5=4=>x=-1
b) 5^x:5^2=125 =>5^x:5^2=5^3 =>5^x=5^3.5^2=5^5 =>5^x=5^5=>x=5
c) (x+1)^2=(x+1)^0 =>x=0 hoặc 1
d) (2+x)+(4+x)+...+(52+x) =780 =>(x+x+...+x) +(2+4+...+52)=780 =>26x+(52+2).26:2=780 =>26x=780-702 =>26x=78=>x=3
d+e) áp dụng công thức ƯC và BC bn nhé. Nếu trình bày ra hơi dài nên bn tự làm nhé.
\(5-\left(1997-2005\right)+1997\)
\(=5-1997+2005+1997\)
\(=2010\)
\(1579-\left(53+1579\right)-\left(-53\right)\)
\(=1579-53-1579+53\)
\(=0\)
\(-48-\left(357-48\right)+300\)
\(=\)\(-48-357+48+300\)
\(=-57\)
\(173-\left(36+173\right)+\left(175-27\right)-175-\left(-36+50\right)\)
\(=173-36-173+175-27-175+36-50\)
\(=-27-50\)
\(=-77\)
\(\)
\(-\left[-171+171+23\right]-\left[172-\left(38+172\right)\right]-49\)
\(=171-171-23-172+38+172-49\)
\(=-23+38-49=-34\)
Bài 1:
\(a,22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
=\(\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{70}{4}+\frac{2}{4}-\frac{5}{4}\)
=\(\frac{67}{4}\)
\(b,1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)
=\(\frac{3}{7}-\frac{2}{3}\)
=\(-\frac{5}{21}\)
\(c,125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)
=\(\frac{5}{4}.\frac{1}{4}:\left(\frac{11}{6}-\frac{8}{5}\right)+1\)
=\(\frac{5}{16}:\frac{7}{30}+1\)
=\(\frac{131}{56}\)
\(d,1,4.\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{1}{5}+\frac{2}{3}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{15}:\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{33}\)
=\(\frac{8}{231}\)
Bài đ làm giống hệt như bài c
Bài 2 :
\(a,\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}=1\\x=\frac{1}{4}:\frac{3}{4}=\frac{1}{3}\end{matrix}\right.\)
Vậy x ∈{1;\(\frac{1}{3}\)}
\(b,\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)
=>\(\frac{19}{15}.x=\frac{19}{10}\)
=>\(x=\frac{19}{10}:\frac{19}{15}=\frac{3}{2}\)
Vậy x ∈ {\(\frac{3}{2}\)}
c,\(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)
=>\(\left[{}\begin{matrix}2.x-\frac{1}{3}=\frac{2}{9}\\2.x-\frac{1}{3}=-\frac{2}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2.x=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\\2.x=-\frac{2}{9}+\frac{1}{3}=\frac{1}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{5}{9}:2=\frac{5}{18}\\x=\frac{1}{9}:2=\frac{1}{18}\end{matrix}\right.\)
Vậy x∈{\(\frac{5}{18};\frac{1}{18}\)}
\(d,x-30\%.x=-1\frac{1}{5}\)
=\(70\%x=-\frac{6}{5}\)
=\(\frac{7}{10}.x=-\frac{6}{5}\)
=>\(x=-\frac{6}{5}:\frac{7}{10}=-\frac{12}{7}\)
Vậy x∈{\(-\frac{12}{7}\)}
Bài 2
a/
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{3}{4}\\\frac{3}{4}.x=\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}\\x=\frac{1}{4}:\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\frac{1}{3}\)
b/ Đặt x làm thừa số chung rồi tính như bình thường
c/ Tương tự câu a
d/ Tương tự câu b
a)
4 . 25 – 12 . 25 + 170 : 10
= (4 . 25) – (12 . 25) + (170 : 10)
= 100 - 300 + 17
= -183
b)
(7 + 33 + 32) . 4 – 3
= (7 + 27 + 9) .4 – 3
= 43 . 4 – 3
= (43 . 4) – 3
= 45
c)
12 : {400 : [500 – (125 + 25 . 7)}
= 12 : {400 : [500 – (125 + 175)}
= 12 : (400: 200)
= 12 : 2
= 6
d)
168 + {[2.(24 + 32) - 2560] : 72}.
= 168 + [2 . (16 + 9) – 1] : 49
= 168 + 49: 49
= 168 + 1
= 167
a)
4 . 25 – 12 . 25 + 170 : 10
= (4 . 25) – (12 . 25) + (170 : 10)
= 100 - 300 + 17
= -183
b)
(7 + 33 + 32) . 4 – 3
= (7 + 27 + 9) .4 – 3
= 43 . 4 – 3
= (43 . 4) – 3
= 45
c)
12 : {400 : [500 – (125 + 25 . 7)}
= 12 : {400 : [500 – (125 + 175)}
= 12 : (400: 200)
= 12 : 2
= 6
d)
168 + {[2.(24 + 32) - 2560] : 72}.
= 168 + [2 . (16 + 9) – 1] : 49
= 168 + 49: 49
= 168 + 1
= 167
a) \(S=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2014\right)+2015\)
\(\Leftrightarrow S=\left(1-2\right)+\left(3-4\right)+....+\left(2013-2014\right)+2015\)
Vì từ 1 đến 2014 có 2014 số hạng => có 1007 cặp => Có 1007 cặp -1 và số 2015
\(\Rightarrow S=\left(-1\right)\cdot1007+2015\)
<=>S=-1007+2015
<=> S=1008
\(E=250:\left\{175-\left[50+3.5^2\left(2014+2015^{40}\right)^0\right]\right\}\)
\(E=250:\left\{175-\left[50+3.5^2.1\right]\right\}\)
\(E=250:\left\{175-\left[50+75\right]\right\}\)
\(E=250:\left\{175-50-75\right\}\)
\(E=250:50=5\)( mk đổi dấu ngoặc xíu nha tại cho nó đỡ bị nhầm )
\(F=\left(6^{27}:6^{25}\right):\left\{780:\left[5.89-\left(125+5.7^2\right)+5.11\right]\right\}\)
\(F=6^2:\left\{780:\left[5.89-370+5.11\right]\right\}\)
\(F=6^2:\left\{780:\left[5.\left(89+11\right)-370\right]\right\}\)
\(F=36:\left\{780:\left[5.100-370\right]\right\}\)
\(F=36:\left\{780:\left[500-370\right]\right\}\)
\(F=36:\left\{780:130\right\}=36:6=6\)
cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^
cám ơn nguyễn quỳnh trang nhiều nha!
^_^