\(\left(3,5-70,84:23+4-3,375.\frac{4}{9}\right):0,78\)
(3,5 − 70,84: 23 + 4 − 3,375 . 4 9) : 0,78
( 3,5 - 70,84 : 23 + 4 - 3,375 . 49 ) : 0,78
= ( 3,5 - 3,08 + 4 - 165,375 ) : 0,78
= ( 0,42 + -161,375 ) : 0,78
= -160,955 : 0,78
= -220,7.
Câu 1:
\(\dfrac{179}{50}-\left(\dfrac{59}{30}+\dfrac{3}{5}\right)\)
Câu 2:
(3,5-70,84:23+4)-3,375.\(\dfrac{4}{9}\):0,78.
\(1)\dfrac{179}{50}-\left(\dfrac{59}{30}+\dfrac{3}{5}\right)\)
\(=\dfrac{179}{50}-\left(\dfrac{118}{60}+\dfrac{36}{60}\right)\)
\(=\dfrac{179}{50}-\dfrac{77}{30}\)
\(=\dfrac{537}{150}-\dfrac{385}{150}\)
\(=\dfrac{152}{150}=\dfrac{76}{75}=1\dfrac{1}{75}\)
Câu 1 : Ta có : 179/50 - ( 59/30 + 3/5 ) .
= 179/50 - ( 59/30 + 18/30 ) .
= 179/50 - 77/30 .
= 537/150 - 385/150.
= 76/75 .
Câu 2 : Ta có : ( 3,5 - 70,84 : 23 + 4 ) - 3,375 . 4/9 : 0,78 .
= ( 3,5 - 3,08 + 4 ) - 27/8 . 4/9 : 39/2 .
= 4,42 - 3/2 . 2/39 .
= 221/50 - 1/13 .
= 2873/650 - 50/650 .
= 2823/50 .
= 56,46 .
Câu 1:
\(\dfrac{179}{50}-\left(\dfrac{59}{30}+\dfrac{18}{30}\right)=\dfrac{179}{50}-\dfrac{77}{30}=\dfrac{537}{150}-\dfrac{385}{150}=\dfrac{152}{150}=\dfrac{76}{75}\)
Câu 2:
\(\left(3,5-70,84:23+4\right)-3,375.\dfrac{4}{9}:0,78=\left(3,5-3,08+4\right)-3,375.\dfrac{4}{9}:0,78=\left(0,42+4\right)-3,375.\dfrac{4}{9}:0,78=4,42-3,375.\dfrac{4}{9}:0,78=4,42-1,5:0,78=4,42-1,9230=2,497\)
tính
3,5 - 70,84 : 23 + 4 -37,375 x 4/9
\(1\times3,5-70,84:23+4-37,375\times\frac{4}{9}.\)
\(=3,5-\frac{77}{25}+4-\frac{299}{18}\)
\(=\left(3,5-\frac{77}{25}\right)+\left(4-\frac{299}{18}\right)\)
\(=\frac{21}{50}+\left(\frac{-277}{18}\right)\)
\(=\frac{-2743}{225}\)
Tính: \(a)\left| { - 3,5} \right|;b)\left| {\frac{{ - 4}}{9}} \right|;c)\left| 0 \right|;d)\left| {2,0(3)} \right|.\)
\(\begin{array}{l}a)\left| { - 3,5} \right| = 3,5;\\b)\left| {\frac{{ - 4}}{9}} \right| = \frac{4}{9};\\c)\left| 0 \right| = 0;\\d)\left| {2,0(3)} \right| = 2,0(3)\end{array}\)
Chú ý:
Nếu \(a \ge 0\) thì \(\left| a \right| = a\)
Nếu \(a < 0\) thì \(\left| a \right| = - a\)
Tính hợp lí:
a) \(\frac{-8}{15}.\left(-30\right).\left(\frac{15}{-8}\right).\frac{9}{10}\)
b) \(2\frac{1}{18}.\frac{23}{24}.\frac{9}{37}.\frac{48}{-15}\)
c) A=\(\frac{-0,8+\frac{4}{7}+\frac{4}{9}}{0,3.\frac{3}{14}-\frac{3}{18}}+\frac{0,23-3,5+\frac{2}{7}}{0,69-10,5+\frac{6}{7}}\)
\(a,\frac{-8}{15}.\left(-30\right).\frac{15}{-8}.\frac{9}{10}\)
\(=-\left(\frac{8}{15}.\frac{15}{8}\right).\left(30.\frac{9}{10}\right)\)
\(=-1.27
=-27\)
\(b,2\frac{1}{18}.\frac{23}{24}.\frac{9}{37}.\frac{48}{-15}\)
\(=\frac{-37.23.9.48}{18.24.37.15}=\frac{23}{15}\)
c, chịu rồi
Tính nhanh:
a)\(\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}};\)
b) \(\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\)
c)\(\left[ {\left( { - \frac{4}{9}} \right) + \frac{3}{5}} \right]:\frac{{13}}{{17}} + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}};\)
d) \(\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\)
a)
\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)
b)
\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)
d)
\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ = - 2\end{array}\)
Tính giá trị biểu thức
\(1.A=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{1}{7}+\left[\frac{-14}{30}\right]\)
\(2.B=\left(\frac{5}{8}-\frac{4}{12}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left[\frac{-3}{2}\right]+\frac{7}{-15}\)
\(3.C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(4.D=\frac{1}{9}-\left[\frac{-5}{23}\right]-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{14}-\frac{7}{30}\)
\(5.E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)+\left(\frac{12}{17}+\frac{5}{18}+\frac{7}{5}\right)\)
\(6.F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{12}{17}-\frac{15}{14}+\frac{1}{4}\right)\)
\(7.G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(8.H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{29}{27}\right)-\frac{2}{27}\)
\(9.K=\frac{1}{16}-\frac{5}{21}+\left(\frac{-1}{16}+\frac{-3}{5}-\frac{-5}{21}\right)+\frac{-2}{5}+\frac{3}{4}\)
\(10.L=\frac{7}{12}+\frac{15}{14}-\left(\frac{14}{22}+\frac{-1}{14}+\frac{5}{21}\right)-\frac{-5}{21}+\frac{3}{5}\)
yutyugubhujyikiu
Bài 1: Thực hiện phép tính:
a)(\(-3\frac{2}{5}+70,84:23-4-3\frac{3}{8}.\frac{-4}{9}\)):75%+25%
b)\(\frac{-16^3.3^5-\left(-8^4\right)\left(-9^2\right)}{\left(-2^2.3\right)^6-\left(-8^4.3^5\right)}-\frac{\left(-25\right)^5.7^3-5^{10}.\left(-7\right)^4}{\left(125.7\right)^3-\left(25.7\right)^3.250}\)
Bài 2: Tìm x:
1-Tìm x, biết:
a) \(3x-\left|4-2x\right|-2+5=-2\left(3-x\right)\)
b) \(2x^2:\left(\frac{-1}{2}-\frac{1}{10}-\frac{3}{70}-\frac{3}{126}-...-\frac{3}{2046}\right)=-11\)
2-Tìm các số nguyên x thỏa mãn:
\(\left(x+5\right)^2=\left[4\left(x-2\right)\right]^3\)
Bài 3: Đội nghi thức của trường Lê Lợi chưa đến 200 em. Khi xếp hàng 5 thì thừa 3 em, khi xếp hàng 7 thiếu 3 em. Khi xếp hàng 9 thiếu 4 em. Tính số học sinh trong đội nghi thức của trường ?
Bài 4: Cho 2 góc kề bù : Góc xOz và góc yOz biết góc xOz bằng\(\frac{1}{3}\)góc yOz. Vẽ điểm A nằm trong góc zOy sao cho AOy = 2 góc AOz. Vẽ Ob là tia phân giác của góc AOy. Chứng minh tia OA là tia phân giác của góc bOz ?
Bài 5: Sau khi đổi chỗ các chữ số của số tự nhiên A đước số B gấp 3 lần . Chứng minh rằng B chia hết cho 27
ai trả lời nhanh nhất và đúng nhất mink cho nha !!!!!
Rút gọn phân thức P=\(\frac{\left(1^4+4\right)\left(5^4 +4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
\(P=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)\(=\frac{\left(1+4\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)...\left(20^2+1\right)\left(\cdot22^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)\left(12^2+1\right)...\left(22^2+1\right)\left(24^2+1\right)}\)
\(=\frac{1+4}{\left(2^2+1\right)\left(24^2+1\right)}=\frac{5}{5\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)
cái bước tách ra bn nhân lại là có kết quả y chang, VD:
\(\left(5^4+4\right)=\left(4^2+1\right)\left(6^2+1\right)=629\)