Cho a+b+c=0 va a^2+b^2+c^2=1
Tinh gia tri cua A= a^4+b^4+c^4=
Tinh gia tri cua bieu thuc a^4+b^4+c^4,biet rang a+b+c=0 va:
a^2+b^2+c^2=2
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=2+2\left(ab+bc+ac\right)\)
=> \(0=2+2\left(ab+bc+ac\right)\)=> \(ab+bc+ca=-1\)
=> \(\left(ab+bc+ac\right)^2=1\)
Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)\)
\(=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+a^2c^2\)
=> \(a^2b^2+b^2c^2+c^2a^2=1\)
Mặt khác : \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
=> \(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=4-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
=> \(a^4+b^4+c^4=4-2=2\)
cho a+b+c=0 va a2+b2+c2=1
tinh gia tri bthuc M=a4+b4+c4
Ta có :
( a + b + c )2 = a2 + b2 + c2 + 2ab + 2 bc+ 2ac = 0
Mà a2 + b 2 + c2 = 1
=> 2ab + 2bc + 2ac = - 1
=> ab + bc + ac = \(\frac{-1}{2}\)
=> ( ab + bc + ac ) 2 = a2b2 + a2c2 + b2c 2 + 2ab2c + 2ac2b + 2a2bc = \(\left(\frac{-1}{2}\right)^2\)=\(\frac{1}{4}\)
=> a2b2 + a2c2 + b2c2 + 2abc ( a + b +c ) = \(\frac{1}{4}\)
mà a + b + c = 0 => 2abc ( a +b +c ) = 0
=> a2b2 + b2c2 + c2a2 = \(\frac{1}{4}\)
Ta có : ( a2 + b2 + c2 )2 = a4 + b4 + c4 + 2 ( a2b2 + b2c2 + c2a2 ) = 1
=> a4 +b4 + c4 + 2. \(\frac{1}{4}\) = 1
=> a4 + b4 + c4 = 1 - \(\frac{1}{2}\)
=> a4 + b4 + c4 = \(\frac{1}{2}\)
Cho a+b+c=0; a2+b2+c2=4. Tinh gia tri cua bieu thuc: A=ab+bc+ca; B= a4+b4+c4
\(\Rightarrow ab+bc+ac=\frac{-\left(a^2+b^2+c^2\right)}{2}=-\frac{4}{2}=-2\)
Ta có ; \(\left(a^2+b^2+c^2\right)^2=16\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=16\)\(\Leftrightarrow a^4+b^4+c^4=16-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
Mặt khác : \(\left(ab+bc+ac\right)^2=4\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=4\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=4\)
\(\Rightarrow a^4+b^4+c^4=16-2.4=8\)
cho a+b+c=0 va a2+b2+c2=14. tinh gia tri bieu thuc B=a4+b4+c4
6.a) tim 2 so x, ybik 7x=2y va x-y=16
b)so sanh a,b,c bik a/b=b/c=c/a
c)tim cac so a,b,cbik a/2=b/3=c/4 va a+2b-c=-20
d)cho x/2=y/5=z/7 tinh gia tri bieu thuc A=x-y+z/x+2y-z
e)cho 3x-2y/4=2z-4x/3=4y-3z/2.CMR x/2=y/3=z/4
f)cho a,b,c la cac so huu ti khac sao choa+b-c/c=a-b+c/b=-a+b+c/a
tinh gia tri bang so cua 1 bieu thuc m=(a+b)(b+c)(c+a)/abc
g)cho x/a=y/b=z/c CMRbz-cy/a=cx-az/b=ay-bx/c
a)Ta có 7x=2y
Suy ra:\(\dfrac{x}{\dfrac{1}{7}}\)=\(\dfrac{y}{\dfrac{1}{2}}\)
Và x-y=16
Áp dụng công thức của dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{\dfrac{1}{7}}\)=\(\dfrac{y}{\dfrac{1}{2}}\)=\(\dfrac{x-y}{\dfrac{1}{7}-\dfrac{1}{2}}\)=\(\dfrac{16}{\dfrac{-5}{14}}\)=\(\dfrac{-224}{5}\)
Từ \(\dfrac{x}{\dfrac{1}{7}}=\dfrac{-224}{5}\)suy ra :x=\(\dfrac{-224}{5}\cdot\dfrac{1}{7}\)=\(-\dfrac{32}{5}\)
\(\dfrac{y}{\dfrac{1}{2}}=-\dfrac{224}{5}\)suy ra:y=\(-\dfrac{224}{5}\cdot\dfrac{1}{2}=-\dfrac{112}{5}\)
c)Ta có :\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
Mà a+2b-c=-20
Suy ra:\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{c}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có:
\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{c}{4}=\dfrac{a+2b-c}{2+6-4}=-\dfrac{20}{4}=-5\)
Từ \(\dfrac{a}{2}=-5,suyra:a=-5\cdot2=-10\)
\(\dfrac{b}{3}=-5,suyra:b=-5\cdot3=-15\)
\(\dfrac{c}{4}=-5,suyra:c=-5\cdot4=-20\)
Vậy a=-10,b=-15,c=-20
Giup minh voi cac ban !!!
Cho a + b + c = 0 va a^2 + b^2 + c^2 = 1 . khi do , gia tri bieu thuc a^4 + b^4 + c^4 = ????
cho a+b+c=0,a2+b2+c2=1.Tinh gia tri bieu thuc: A=a4+b4+c4
Ta có a + b + c = 0
<=> (a + b + c)2 = 0
<=> a2 + b2 + c2 + 2(ab + bc + ca) = 0
<=> ab + bc + ca = \(-\frac{1}{2}\)
=> \(\left(ab+bc+ca\right)^2=\frac{1}{4}\)
<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2a^2bc+2abc^2=\frac{1}{4}\)
<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\frac{1}{4}\)
Lại có a2 + b2 + c2 = 1
=> (a2 + b2 + c2)2 = 1
<= > a4 + b4 + c4 + 2[(ab)2 + (bc)2 + (ca)2] = 1
<=> \(a^4+b^4+c^4+2.\frac{1}{4}=1\)
<=> \(a^4+b^4+c^4=\frac{1}{2}\)
Từ a + b + c = 0 => ( a + b + c )2 = 0 <=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
<=> ab + bc + ca = -1/2 => ( ab + bc + ca )2 = 1/4
<=> a2b2 + b2c2 + c2a2 + 2ab2c + 2bc2a + 2a2bc = 1/4
<=> a2b2 + b2c2 + c2a2 + 2abc( a + b + c ) = 1/4
<=> a2b2 + b2c2 + c2a2 = 1/4 ( vì a + b + c = 0 )
Từ a2 + b2 + c2 = 1 => ( a2 + b2 + c2 )2 = 1 <=> a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2 = 1
<=> a4 + b4 + c4 + 2( a2b2 + b2c2 + c2a2 ) = 1
<=> a4 + b4 + c4 + 1/2 = 1 <=> a4 + b4 + c4 = 1/2
Vậy A = 1/2
bai 1
a = (3x / 2x + 4 ) + (x +3 /x ^ 2 - 4 )
a . tim x de gia tri phan thuc a duoc xac dinh
b. rut gon a
c. tinh gia trin cua a khi x bang -3
d . tim gia tri cua x de phan thuc co gia tri bang 2
bai 1
a = (3x / 2x + 4 ) + (x +3 /x ^ 2 - 4 )
a . tim x de gia tri phan thuc a duoc xac dinh
b. rut gon a
c. tinh gia trin cua a khi x bang -3
d . tim gia tri cua x de phan thuc co gia tri bang 2
a: ĐKXĐ: x<>2; x<>-2
b: \(A=\dfrac{3x\left(x-2\right)+2x+6}{2\left(x-2\right)\left(x+2\right)}=\dfrac{3x^2-6x+2x+6}{2\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3x^2+4x+6}{2\left(x-2\right)\left(x+2\right)}\)
c: Khi x=-3 thì \(A=\dfrac{3\cdot\left(-3\right)^2-4\cdot3+6}{2\left(-3-2\right)\left(-3+2\right)}=\dfrac{21}{10}\)