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Ta có:
\(A=\frac{1}{358}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right).2.\frac{1}{297}-7.\frac{1}{358}-\frac{3}{297}.\frac{1}{358}\)
\(=\frac{1}{358}.\left(7+\frac{1}{297}-7-\frac{3}{297}\right)-\left(4-\frac{1}{358}\right).\frac{2}{297}\)
\(=\frac{1}{358}.\left(-\frac{2}{297}\right)-\frac{2}{297}.\left(4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(\frac{1}{358}+4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(-4\right)\)
\(=\frac{8}{297}\)
Vậy giá trị biểu thức A là \(\frac{8}{297}\)
nhìn căng nhể :))
a) ( x - 1 )( x - 3 )( x + 5 )( x + 7 ) - 297 = 0
<=> [ ( x - 1 )( x + 5 ) ][ ( x - 3 )( x + 7 ) ] - 297 = 0
<=> ( x2 + 4x - 5 )( x2 + 4x - 21 ) - 297 = 0
Đặt t = x2 + 4x - 5
pt <=> t( t - 16 ) - 297 = 0
<=> t2 - 16t - 297 = 0
<=> t2 - 27t + 11t - 297 = 0
<=> t( t - 27 ) + 11( t - 27 ) = 0
<=> ( t - 27 )( t + 11 ) = 0
<=> ( x2 + 4x - 5 - 27 )( x2 + 4x - 5 + 11 ) = 0
<=> ( x2 + 4x - 32 )( x2 + 4x + 6 ) = 0
<=> ( x2 - 4x + 8x - 32 )( x2 + 4x + 6 ) = 0
<=> [ x( x - 4 ) + 8( x - 4 ) ]( x2 + 4x + 6 ) = 0
<=> ( x - 4 )( x + 8 )( x2 + 4x + 6 ) = 0
Đến đây dễ rồi :)
- a/ [x/x^2-4 -2(x+2)/x^2-4 +x-2/x^2-4]:[x^2-4/x+2 +10-x^2/x+2] =(x-2x-4+x-2/x^2-4):(x^2-4+10-x^2/x+2) = - 6/x^2-4 nhân với x+2/x^2-4+10-x^2= - 6/(x+2)(x-2) nhân với x+2/6= - 1/x-2.
c/đễ A<0 <=> -1/X-2 <0 <=> x-2<0 <=>x<2
\(x^3+y^3+z^3=3xyz\)
\(x^3+y^3+z^3-3xyz=0\)
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
\(x^2+y^2+z^2-xy-xz-yz=0\left(x+y+z\ne0\right)\)
\(2\times\left(x^2+y^2+z^2-xy-xz-yz\right)=0\times2\)
\(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
\(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
\(\left[\begin{array}{nghiempt}x-y=0\\x-z=0\\y-z=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=y\\x=z\\y=z\end{array}\right.\)
x = y = z
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{x}{z}\right)\)
\(=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
\(=2^3\)
\(=8\)
\(A=\dfrac{1}{358}.\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=7.\dfrac{1}{358}+\dfrac{1}{297}.\dfrac{1}{358}-4.2.\dfrac{1}{297}+2.\dfrac{1}{297}.\dfrac{1}{358}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=\left(7.\dfrac{1}{358}-7.\dfrac{1}{358}\right)+\left(\dfrac{1}{297}.\dfrac{1}{358}+2.\dfrac{1}{297}.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\right)-4.2.\dfrac{1}{297}\)
\(A=0+0+\dfrac{-8}{297}\)
\(A=\dfrac{-8}{297}\)
Chúc bạn học tốt!!!
A= \(\dfrac{1}{358}\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\)
A= \(\dfrac{7}{358}+\dfrac{1}{358.297}-\dfrac{8}{297}+\dfrac{2}{358.297}-\dfrac{7}{358}-\dfrac{3}{358.297}\)
A= \(-\dfrac{8}{297}\)