\(\left(a1^2+a2^2\right)\cdot\left(b1^2+b2^2\right)>\left(a1b1+a2b2\right)\)
Những câu hỏi liên quan
Nếu a1b1=a2b2 thì: A. a1/a2=b1/b2 B. a1/a2=b2/b1 C. a1/b2=a2/b1 D. a1/b2=b1/a2
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Cho các số thực dương a1,b1,c1,a2,b2,c2 thỏa mãn điều kiện \(\frac{a1}{a2}=\frac{b1}{b2}=\frac{c1}{c2}\).CMR \(\sqrt{\left(a1+b1+c1\right)\left(a2+b2+c2\right)}=\sqrt{a1a2}+\sqrt{b1b2}+\sqrt{c1c2}\)
a1/a2 = b1/b2 = c1/c2 = k
a1=k.a2, b1=k.b2, c1=k.c2
Biểu thức trở thành
√(k.a2 + k.b2 + k.c2).(a2 + b2 + c2)= √k.a2.a2 + √k.b2.b2 + √k.c2.c2
√k.(a2+b2+c2)2 = a2. √k + b2. √k + c2. √k
(a2+b2+c2). √k = (a2+b2+c2). √k (hiển nhiên đúng)
Suy ra điều phải chứng minh
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Cho các giá trị ô tính như sau:
A1=10; A2=30; A3=20.
B1=40; B2=20; B3=50; B4=100.
a) =AVERAGE\(\left[Sum\left(A1,A2,A3\right)\right]\)
b)=SUM(A1, B1:B4)
c)=MAX(A1, B1,B4)
d)=SUM\(\left[AVERAGE\left(B1:B4\right)\right]\)
Câu 4: Cho bảng tính như hình bên:
|
Hãy điền kết quả vào bảng sau:
Công thức tại ô D1 | Kết quả |
=SUM(A1:C3,1) |
|
=AVERAGE(A2:C2) |
|
=MIN(A2:C2,4) |
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=MAX(A3:C3) |
|
=AVERAGE(A3:C3) + MAX(A2:C2) |
|
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Bài 1Cho frac{a+b+c}{a+b-c}frac{a-b+c}{a-b-c}left(bne0right)Chững minh c0Bài 2Cho tỉ lệ thức frac{a+b}{b+c}frac{c+d}{d+a}Chững minh a + b+ c+ d 0Bài 3Cho frac{cx-az}{b}frac{ay-bx}{c}frac{bz-cy}{a}Chững mình rằng frac{x}{a}frac{y}{b}frac{z}{c}Bài 4Cho a + b c + d và a^2+b^2+c^2c^2+d^2left(a,b,c,dne0right)Chững minh rằng 4 số a,b, c, d lập thành 1 tỉ lệ thứcBài 5Cho left(x1P-y1Qright)^{2n}+left(x2P+y2Qright)^{2m}+...+left(xkP-ykQright)^{2k}le0left(n,m,...,kinℕ^∗;P,Qne0right)Chứng minh rằng frac{...
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Bài 1
Cho \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}\left(b\ne0\right)\)
Chững minh c=0
Bài 2
Cho tỉ lệ thức \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)
Chững minh a + b+ c+ d = 0
Bài 3
Cho \(\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bz-cy}{a}\)
Chững mình rằng \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Bài 4
Cho a + b = c + d và \(a^2+b^2+c^2=c^2+d^2\left(a,b,c,d\ne0\right)\)
Chững minh rằng 4 số a,b, c, d lập thành 1 tỉ lệ thức
Bài 5
Cho \(\left(x1P-y1Q\right)^{2n}+\left(x2P+y2Q\right)^{2m}+...+\left(xkP-ykQ\right)^{2k}\le0\left(n,m,...,k\inℕ^∗;P,Q\ne0\right)\)
Chứng minh rằng \(\frac{x1+x2+x3+...+xk}{y1+y2+y3+...+yk}\)
Bài 6
Biết rằng \(\hept{\begin{cases}a1^2+a2^2+a3^2=P^2\\b1^2+b2^2+b3^2=Q^2\end{cases}}\) và \(a1\cdot b1+a2\cdot b2+a3\cdot b3=P\cdot Q\)
Chứng minh \(\frac{a1}{b1}=\frac{a2}{b2}=\frac{a3}{b3}=\frac{P}{Q}\)
Bài 7
Cho 4 số a, b, c, d khác 0 thảo mãn \(\left(ad+bc\right)^2=4abcd\)
Chững minh rằng 4 số a, b, c ,d có thê rlaapj thành 1 tỉ lệ thức
Bài 8
Cho các số a, b, c thảo mãn \(\frac{a}{2010}=\frac{b}{2011}=\frac{c}{2012}\)
a. Tính \(M=\frac{2a-3b+c}{2c-3b}\)
b. Chứng minh rằng \(a\cdot\left(a-b\right)\cdot\left(b-c\right)=\left(a-c\right)^2\)
Cho a,b > 0 và a2+b2=1. Tìm GTNN của biểu thức sau :
P = \(\left(2+a\right)\left(1+\dfrac{1}{b}\right)+\left(2+b\right)\left(1+\dfrac{1}{a}\right)\)
\(P=2+\dfrac{2}{b}+a+\dfrac{a}{b}+2+\dfrac{2}{a}+b+\dfrac{b}{a}=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(a+\dfrac{1}{2a}\right)+\left(b+\dfrac{1}{2b}\right)+\left(\dfrac{3}{2a}+\dfrac{3}{2b}\right)+4\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{a.\dfrac{1}{2a}}+2\sqrt{b.\dfrac{1}{2b}}+2\sqrt{\dfrac{3}{2a}.\dfrac{3}{2b}}+4=6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\)
Ta lại có: \(a^2+b^2\ge2\sqrt{a^2.b^2}=2ab\left(BĐT.Cauchy\right)\Rightarrow2\left(a^2+b^2\right)\ge4ab\Rightarrow\sqrt{ab}\le\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}=\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow P\ge6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\ge6+2\sqrt{2}+\dfrac{3}{\dfrac{\sqrt{2}}{2}}=6+5\sqrt{2}\)
\(minP=6+5\sqrt{2}\Leftrightarrow a=b=\dfrac{\sqrt{2}}{2}\)
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\(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)\cdot\cdot\cdot\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)\cdot\cdot\cdot\left(99\times102+2\right)}=...\)
Tính:-3^2+left{-54:left[-2^8+7right]cdotleft(-2right)^2right}Tính hợp lí :31cdotleft(-18right)+31cdotleft(-81right)-31left(-12right)cdot47+left(-12right)cdot52+left(-12right)13cdotleft(23+22right)-3cdotleft(17+28right)-48+48cdotleft(-78right)+48cdotleft(-21right)
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Tính:
\(-3^2+\left\{-54:\left[-2^8+7\right]\cdot\left(-2\right)^2\right\}\)
Tính hợp lí :
\(31\cdot\left(-18\right)+31\cdot\left(-81\right)-31\)
\(\left(-12\right)\cdot47+\left(-12\right)\cdot52+\left(-12\right)\)
\(13\cdot\left(23+22\right)-3\cdot\left(17+28\right)\)
\(-48+48\cdot\left(-78\right)+48\cdot\left(-21\right)\)
`#3107.101107`
`-3^2 + {-54 \div [-2^8 + 7] * (-2)^2}`
`= -9 + [-54 \div (-256 + 7) * 4]`
`= -9 + [-54 \div (-249) * 4]`
`= -9 + (18/83 * 4)`
`= -9 + 72/83`
`= -675/83`
______
`31 * (-18) + 31 * (-81) - 31`
`= 31 * (-18 - 81 - 1)`
`= 31 * (-100)`
`= -3100`
___
`(-12) * 47 + (-12) * 52 + (-12)`
`= (-12) * (47 + 52 + 1)`
`= (-12) * 100`
`= -1200`
___
`13 * (23 + 22) - 3 * (17 + 28)`
`= 13 * 45 - 3 * 45`
`= 45 * (13 - 3)`
`= 45 * 10`
`= 450`
____
`-48 + 48 * (-78) + 48 * (-21)`
`= 48 * (-1 - 78 - 21)`
`= 48 * (-100)`
`= -4800`
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rút gọn biểu thức sau bằng cách nhanh nhấtA left(a^2+b^2-c^2right)^2-left(a^2-b^2+c^2right)^2-4a^2b^2B left(3x^3+3x+1right)cdotleft(3x^3-3x+1right)-left(3x^3+1right)^2C left(2-6xright)^2+left(2-5xright)^2+2cdotleft(6x-2right)cdotleft(2-5xright)D 5cdotleft(3x-1right)^2+4cdotleft(5x+1right)^2-12cdotleft(5x-2right)left(5x+2right)E left(3x-1right)^2+left(2x+4right)cdotleft(1-3xright)+left(x+2right)^2G left(x-1right)^3+4cdotleft(x+1right)cdotleft(1-xright)+3cdotleft(x-1right)cdotleft(x^2+x+1rig...
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rút gọn biểu thức sau bằng cách nhanh nhất
A = \(\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
B = \(\left(3x^3+3x+1\right)\cdot\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
C = \(\left(2-6x\right)^2+\left(2-5x\right)^2+2\cdot\left(6x-2\right)\cdot\left(2-5x\right)\)
D = \(5\cdot\left(3x-1\right)^2+4\cdot\left(5x+1\right)^2-12\cdot\left(5x-2\right)\left(5x+2\right)\)
E = \(\left(3x-1\right)^2+\left(2x+4\right)\cdot\left(1-3x\right)+\left(x+2\right)^2\)
G = \(\left(x-1\right)^3+4\cdot\left(x+1\right)\cdot\left(1-x\right)+3\cdot\left(x-1\right)\cdot\left(x^2+x+1\right)\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
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Thu gọn biểu thức sau :
a) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot\left(\frac{1}{16}+1\right)\cdot\cdot\cdot\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
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a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
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