14/20< ?/26<15/20
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a)\(A=^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\)
=> \(A^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)
\(=20+14\sqrt{2}+20-14\sqrt{2}\)
\(+3\left(\text{}^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\right)\left(^3\sqrt{20+14\sqrt{2}}.^3\sqrt{20-14\sqrt{2}}\right)\)
\(=40+3A.^3\sqrt{\left(20+14\sqrt{2}\right)\left(20+14\sqrt{2}\right)}\)
\(\Rightarrow A^3=40+3.A.2\)
=> \(A^3-6A-40=0\)
<=> \(A^3-16A+10A-40=0\)
<=> \(A\left(A-4\right)\left(A+4\right)+10\left(A-4\right)=0\)
<=> \(\left(A-4\right)\left(A^2+4A+10\right)=0\)
<=> A = 4 ( vì \(A^2+4A+10=\left(A+2\right)^2+6>0\))
Vậy A = 4.
b/ \(B=^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\)
=> \(B^3=\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right)^3\)
\(=26+15\sqrt{3}-26+15\sqrt{3}\)
\(-3\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right).^3\sqrt{26+15\sqrt{3}}.^3\sqrt{26-15\sqrt{3}}\)
\(=30\sqrt{3}-3B.1\)
=> \(B^3+3B-30\sqrt{3}=0\)
<=> \(B^3-12B+15B-30\sqrt{3}=0\)
<=> \(B\left(B-2\sqrt{3}\right)\left(B+2\sqrt{3}\right)+15\left(B-2\sqrt{3}\right)=0\)
<=> \(\left(B-2\sqrt{3}\right)\left(B^2+2\sqrt{3}B+15\right)=0\)
<=> \(B-2\sqrt{3}=0\)( vì \(B^2+2\sqrt{3}B+15=\left(B+\sqrt{3}\right)^2+12>0\))
<=> \(B=2\sqrt{3}\)
\(20+\left(27+13\right)+\left(14+26\right)\)
\(=20+40+40\)
\(=100\)
Lớp 3 chưa học SSH nên giải theo cách này:
(30 + 2) + (4 + 28) + ( 6 + 26) + (8 + 24) + (10 + 12) + (14 + 18) + 16
= 32 + 32 + 32 + 32 + 32 + 32 + 16
= 32 x 6 + 16
= 208
2+4+6+8+10+12+14+16+18+20+22+24+26+30
=(2+8)+(4+6)+10+(12+18)+(14+16)+(24+26)+20+22+30
=10+10+10+30+30+50+20+22+30
=(30+30+30+10)+(10+10+50+20)+22
=100+90+22
=190+22
=212
mk tra loi truoc h mk nha
\(\dfrac{14}{20}\) < \(\dfrac{?}{26}\) < \(\dfrac{15}{20}\)
\(\dfrac{14\times13}{20\times13}\) < \(\dfrac{x\times10}{26\times10}\) < \(\dfrac{15\times13}{20\times13}\)
14 \(\times\) 13 < \(x\times10\) < 15 \(\times\) 13
182 < \(x\) \(\times\) 10 < 195
\(\dfrac{182}{10}\) < \(x\) < \(\dfrac{195}{10}\)
18,2 < \(x\) < 19,5
Vì \(x\) là số tự nhiên nên \(x\) = 19
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