1.[(3√2)-(2√3)]^2-[(3√2)+2√3)]^22.√3÷√2×(14/3-√2)-(√24+√12)3.化简:[(a-2√ab+b)/√a-√b]-(√a+√b)4.已知a-b=2+√3,b-c=2-√3,求代数式a^2+b^2+c^2-ab-bc-ca的值.5.设n是自然数,且x>1,比较(√n+1)-√n和√n-(√
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![1.[(3√2)-(2√3)]^2-[(3√2)+2√3)]^22.√3÷√2×(14/3-√2)-(√24+√12)3.化简:[(a-2√ab+b)/√a-√b]-(√a+√b)4.已知a-b=2+√3,b-c=2-√3,求代数式a^2+b^2+c^2-ab-bc-ca的值.5.设n是自然数,且x>1,比较(√n+1)-√n和√n-(√](/uploads/image/z/5483353-49-3.jpg?t=1.%5B%283%E2%88%9A2%29-%282%E2%88%9A3%29%5D%5E2-%5B%283%E2%88%9A2%29%2B2%E2%88%9A3%29%5D%5E22.%E2%88%9A3%C3%B7%E2%88%9A2%C3%97%2814%2F3-%E2%88%9A2%29-%28%E2%88%9A24%2B%E2%88%9A12%293.%E5%8C%96%E7%AE%80%3A%5B%28a-2%E2%88%9Aab%2Bb%29%2F%E2%88%9Aa-%E2%88%9Ab%5D-%28%E2%88%9Aa%2B%E2%88%9Ab%294.%E5%B7%B2%E7%9F%A5a-b%3D2%2B%E2%88%9A3%2Cb-c%3D2-%E2%88%9A3%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8Fa%5E2%2Bb%5E2%2Bc%5E2-ab-bc-ca%E7%9A%84%E5%80%BC.5.%E8%AE%BEn%E6%98%AF%E8%87%AA%E7%84%B6%E6%95%B0%2C%E4%B8%94x%3E1%2C%E6%AF%94%E8%BE%83%28%E2%88%9An%2B1%29-%E2%88%9An%E5%92%8C%E2%88%9An-%28%E2%88%9A)
1.[(3√2)-(2√3)]^2-[(3√2)+2√3)]^22.√3÷√2×(14/3-√2)-(√24+√12)3.化简:[(a-2√ab+b)/√a-√b]-(√a+√b)4.已知a-b=2+√3,b-c=2-√3,求代数式a^2+b^2+c^2-ab-bc-ca的值.5.设n是自然数,且x>1,比较(√n+1)-√n和√n-(√
1.[(3√2)-(2√3)]^2-[(3√2)+2√3)]^2
2.√3÷√2×(14/3-√2)-(√24+√12)
3.化简:[(a-2√ab+b)/√a-√b]-(√a+√b)
4.已知a-b=2+√3,b-c=2-√3,求代数式a^2+b^2+c^2-ab-bc-ca的值.
5.设n是自然数,且x>1,比较(√n+1)-√n和√n-(√n-1)的大小
1.[(3√2)-(2√3)]^2-[(3√2)+2√3)]^22.√3÷√2×(14/3-√2)-(√24+√12)3.化简:[(a-2√ab+b)/√a-√b]-(√a+√b)4.已知a-b=2+√3,b-c=2-√3,求代数式a^2+b^2+c^2-ab-bc-ca的值.5.设n是自然数,且x>1,比较(√n+1)-√n和√n-(√
1.[(3√2)-(2√3)]^2-[(3√2)+2√3)]^2
=-4(3√2)(2√3)
=-24√6
2.√3÷√2×(14/(3-√2)-(√24+√12)
=√(3/2)*14(3+√2)/[(3-√2)(3+√2)]-2√6-2√3
=√6/2*(6+2√2)-2√6-2√3
=3√6+2√3-2√6-2√3
=√6
3.化简:[(a-2√ab+b)/√a-√b]-(√a+√b)
=(√a-√b)^2/(√a-√b)-(√a+√b)
=(√a-√b)-(√a+√b)
=-2√b
4.已知a-b=2+√3,b-c=2-√3,求代数式
a^2+b^2+c^2-ab-bc-ca的值.
a^2+b^2+c^2-ab-bc-ca
=1/2[(a-b)^2+(b-c)^2+(c-a)^2]
=1/2[(a-b)^2+(b-c)^2+(a-b+b-c))^2]
=1/2*[(2+√3)^2+(2-√3)^2+(2+√3+2-√3)^2]
=1/2*(8+6+16)
=15
5.设n是自然数,且x>1,比较(√n+1)-√n和√n-(√n-1)的大小
(√n+1)-√n=1/[√(n+1)+√n]
√n-(√n-1)=1/[√n+√(n-1)]
因为[√(n+1)+√n]>[√n+√(n-1)]
所以1/[√(n+1)+√n]
1. [(3√2)-(2√3)]^2-[(3√2)+2√3)]^2
=[(3√2)-(2√3)+3√2)+(2√3)][3√2)-(2√3)-(3√2)-(2√3)]^]
=6√2[-4√3)]
=-24√6
2. √3÷√2×(14/3-√2)-(√24+√12)
3. 化简:[(a-2√ab+b)/√a-√b]-(√a+√b)
4. 已知a...
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1. [(3√2)-(2√3)]^2-[(3√2)+2√3)]^2
=[(3√2)-(2√3)+3√2)+(2√3)][3√2)-(2√3)-(3√2)-(2√3)]^]
=6√2[-4√3)]
=-24√6
2. √3÷√2×(14/3-√2)-(√24+√12)
3. 化简:[(a-2√ab+b)/√a-√b]-(√a+√b)
4. 已知a-b=2+√3,b-c=2-√3,求代数式a^2+b^2+c^2-ab-bc-ca的值.
5. 设n是自然数,且x>1,比较(√n+1)-√n和√n-(√n-1)的大小
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