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\(\left(45\frac{1}{2}-2\frac{3}{8}\right)-\left(5\frac{5}{6}+6\frac{3}{4}\right)+\left(10\frac{2}{3}-5\frac{5}{8}\right)\)
\(=\left(\frac{91}{2}-\frac{19}{8}\right)-\left(\frac{35}{6}+\frac{27}{4}\right)+\left(\frac{32}{3}-\frac{45}{8}\right)\)
\(=\frac{345}{8}-\frac{151}{12}+\frac{121}{24}\)
\(=\frac{1035}{24}-\frac{302}{24}+\frac{121}{24}\)
\(=\frac{854}{24}\)\(=\frac{427}{12}\)
C1: Gọi ba số lần lượt là a,b,c
Ta có: \(b=\frac{4}{3}a\Rightarrow4a=3b\Rightarrow\frac{a}{3}=\frac{b}{4}\Rightarrow\frac{a}{9}=\frac{b}{12}\)
\(b=\frac{3}{4}c\Rightarrow4b=3c\Rightarrow\frac{b}{3}=\frac{c}{4}\Rightarrow\frac{b}{12}=\frac{c}{16}\)
\(\Rightarrow\frac{a}{9}=\frac{b}{12}=\frac{c}{16}\Rightarrow\frac{a^2}{81}=\frac{b^2}{144}=\frac{c^2}{256}=\frac{a^2+b^2+c^2}{81+144+256}=\frac{481}{481}=1\)
=> \(\frac{a^2}{81}=1\Rightarrow a^2=81\Rightarrow a=\pm9\)
\(\frac{b^2}{144}=1\Rightarrow b^2=144\Rightarrow b=\pm12\)
\(\frac{c^2}{256}=1\Rightarrow c^2=256\Rightarrow c=\pm16\)
C2: Làm tiếp phần c1
Đặt \(\frac{a}{9}=\frac{b}{12}=\frac{c}{16}=k\Rightarrow a=9k,b=12k,c=16k\)
Ta có: a2 + b2 + c2 = 481
=> (9k)2 + (12k)2 + (16k)2 = 481
=> 81k2 + 144k2 + 256k2 = 481
=> k2(81 + 144 + 256) = 481
=> 481k2 = 481
=> k2 = 1
=> k = \(\pm\)1
Với k = 1 => a = 9, b = 12, c = 16
Với k = -1 => a = -9, b = -12, c = -16
Vậy...
a, 0,(37)+0,(62)=0,(99)
Theo quy ước làm tròn số ta dược :
0,\left(99\right)\approx10,(99)≈1 (đpcm)
b) Làm tương tự câu a) ta có :
0,\left(33\right).3=0,\left(99\right)\approx10,(33).3=0,(99)≈1 (đpcm)
a) Ta có:
0,\left(37\right)=\frac{37}{99}0,(37)=9937 ; 0,\left(62\right)=\frac{62}{99}0,(62)=9962
=> 0,\left(37\right)+0,\left(62\right)=\frac{37}{99}+\frac{62}{99}=\frac{99}{99}=10,(37)+0,(62)=9937+9962=9999=1
b) Ta có:
0,\left(33\right)=\frac{33}{99}0,(33)=9933
=> 0,\left(33\right).3=\frac{33}{99}.3=\frac{1}{3}.3=10,(33).3=9933.3=31.3=1
\(A=\)bn ghi lại đề nha mình lười
\(=\left(\sqrt{5+\sqrt{21}}\right)^2\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
\(=\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{5-\sqrt{21}}\right)\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{14}-\sqrt{6}\right)\)
\(=\left(\sqrt{\left(5^2-21\right)}\right)\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{14}-\sqrt{6}\right)\)
\(=2.\left(\sqrt{5+\sqrt{21}}\right)\sqrt{2}.\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2.\left(\sqrt{10+2\sqrt{21}}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2.\left(\sqrt{7+2\sqrt{21}+3}\right) \left(\sqrt{7}-\sqrt{3}\right)\)
\(=2.\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2.\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=2.\left(7-3\right)=2.4=8\)
tíck mình nha bn thanks nhìu !!!!!!!!!
