a)\(\left(-24\right).17+\left(-35\right).24+\left(-24\right).48\)
b)\(145+\left(27-145\right)-100+27\)
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4.Tinha,frac{4^{30}.3^{43}}{2^{53}.27^{15}} b,frac{left(-27right)^{10}.16^{25}}{6^{30}.left(-32right)^{15}}c,left(frac{-2}{17}right)^3.left(frac{-17}{35}right)^3.left(frac{35}{3}right)^3d,frac{3^{31}.25^8-9^{14}.5^{15}}{27^{10}.5^{15}-3^{30}.5^{17}}e,2^{^2}+4^{^2}+6^2+..........+24^2left(1^2+2^2+3^2+....+12^2650right)
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4.Tinh
a,\(\frac{4^{30}.3^{43}}{2^{53}.27^{15}}\)
b,\(\frac{\left(-27\right)^{10}.16^{25}}{6^{30}.\left(-32\right)^{15}}\)
c,\(\left(\frac{-2}{17}\right)^3.\left(\frac{-17}{35}\right)^3.\left(\frac{35}{3}\right)^3\)
d,\(\frac{3^{31}.25^8-9^{14}.5^{15}}{27^{10}.5^{15}-3^{30}.5^{17}}\)
e,\(2^{^2}+4^{^2}+6^2+..........+24^2\left(1^2+2^2+3^2+....+12^2=650\right)\)
\(\frac{6+a}{9-a^2}+...=\frac{6}{9-a^2}+...+\frac{a^2}{9a-a^3}\ge\frac{54}{27-a^2-b^2-c^2}+\frac{\left(a+b+c\right)^2}{9\left(a+b+c\right)-\left(a^3+b^3+c^3\right)}\)
\(\ge\frac{54}{27-2\left(a+b+c\right)+3}+\frac{9}{27-3\left(a+b+c\right)+6}=\frac{54}{24}+\frac{9}{24}=\frac{21}{8}\)
đây là toán đâu phải văn. bạn bị say rượu à
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\(A=6.25-\left(61-17\right).12+\left(27+24\right):17\)
\(B=\left(-126\right)-\left(4^2-5^2\right)+870:29\)
\(7^2.5-\left(2x+1\right)=630:9\)\(\)
\(\left(12-4x\right)+120:23=17\)
Tính:a)frac{2}{{15}} + left( {frac{{ - 5}}{{24}}} right) b) left( {frac{{ - 5}}{9}} right) - left( { - frac{7}{{27}}} right); c)left( { - frac{7}{{12}}} right) + 0,75d)left( {frac{{ - 5}}{9}} right) - 1,25 e)0,34.frac{{ - 5}}{{17}} g) frac{4}{9}:left( { - frac{8}{{15}}} right);h)left( {1frac{2}{3}} right):left( {2frac{1}{2}} right) i) frac{2}{5}.left( { - 1.25} right) k) left( {frac{{ - 3}}{5}} right).left( {frac{{15}}{...
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Tính:
a)\(\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right)\)
b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right);\)
c)\(\left( { - \frac{7}{{12}}} \right) + 0,75\)
d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25\)
e)\(0,34.\frac{{ - 5}}{{17}}\)
g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right);\)
h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right)\)
i) \(\frac{2}{5}.\left( { - 1.25} \right)\)
k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9}\)
\(a)\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ - 25}}{{120}}} \right) = \frac{{ - 9}}{{120}} = \frac{{ - 3}}{{40}}\)
b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right) = \left( {\frac{{ - 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ - 8}}{{27}}\)
c)\(\left( { - \frac{7}{{12}}} \right) + 0,75 = \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \\= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \\= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)
d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25 =\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\\ = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)
e)\(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ - 5}}{{17}} = \frac{{ - 1}}{{10}}\)
g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { - \frac{{15}}{8}} \right) = \frac{{ - 5}}{6}\)
h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)
i) \(\frac{2}{5}.\left( { - 1,25} \right) = \frac{2}{5}.\frac{{ - 125}}{100} = \frac{2}{5}.\frac{{ - 5}}{4} = \frac{{ - 1}}{2}\)
k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} = \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\\ = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)
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13 tháng 11 2023 lúc 21:17
so sánh: \(A=26^2-24^2\) và \(B=27^2-25^2\)
tìm x, biết:
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
Bài 1:
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)
\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)
=>A<B
Bài 2:
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
=>4x+13=11
=>4x=-2
=>\(x=-\dfrac{1}{2}\)
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Bài 1:a)left|x-2right|+left|1-dfrac{x}{2}right|0 b)left(x-dfrac{1}{3}right)^3dfrac{-8}{27}c)dfrac{x^2}{6}dfrac{24}{25} c)dfrac{x-1}{x+5}dfrac{6}{7}Giúp mik làm bài này với ạ mik đang cần gấp tối nay mik phải nộp rồi mong mn giúp đỡ mik. Mik cảm ơn mn
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Bài 1:
a)\(\left|x-2\right|\)+\(\left|1-\dfrac{x}{2}\right|\)=0 b)\(\left(x-\dfrac{1}{3}\right)^3\)=\(\dfrac{-8}{27}\)
c)\(\dfrac{x^2}{6}\)=\(\dfrac{24}{25}\) c)\(\dfrac{x-1}{x+5}\)=\(\dfrac{6}{7}\)
Giúp mik làm bài này với ạ mik đang cần gấp tối nay mik phải nộp rồi mong mn giúp đỡ mik. Mik cảm ơn mn
tìm x biết :
a) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
c) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
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\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
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Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)
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Xem thêm câu trả lời
Đưa thừa số ra ngoài dấu căn
a.
\(\left(\sqrt{28}-5\sqrt{35}+7\sqrt{112}\right)2\sqrt{7}\)
b. \(\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
a) \(\left(\sqrt{28}-5\sqrt{35}+7\sqrt{112}\right)2\sqrt{7}=2\sqrt{196}-10\sqrt{245}+14\sqrt{784}\)
\(=28-10\sqrt{49.5}+392=420-70\sqrt{5}\)
b) \(\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}=\sqrt{144}-3\sqrt{48}+5\sqrt{16}+4\sqrt{9.3}\)
\(=12-3\sqrt{16.3}+20+12\sqrt{3}=32-12\sqrt{3}+12\sqrt{3}=32\)
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\(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
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