mo ta thuat toan sau tinh gia tri bieu thuc p=b^3+(a-b)+c*d-4
cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
CHO bieu thuc
B=17,58*43+57*17,58/293*A
a)tim gia tri cua bieu thuc Bkhi a=2
b)tinh gia tri a khi B=2
c)tim gia tri so tu nhien ad biet bieu thuc B co gia tri lon nhat ,gia tri lon do la bao nhieu
Cho phan thuc B=(3\y+3)+(1\y-3)-(18\9-y2)
a)Tim dieu kien cua y de gia tri cua bieu thuc B duoc xac dinh
b)Rut gon bieu thuc B
c)Tinh gia tri cua B de B co gia tri nguyen
cho a R b =a*b/ a+b . Tinh gia tri bieu thuc cua : 13R7 - 12r4 . TINH gia tri cua bieu thuc tren ( 30 giay )
\(\frac{31}{20}\)đáp số đúng 1000000000000000000000000000000000000000000000000000% đó nha
h and kb nhé
Cho bieu thuc A = \(^{x2+4x+3}\)
a Tinh gia tri bieu thuc tai x= \(\frac{-1}{2}\)
b Tinh gia tri x de bieu thuc A bang 0
a. Tại x=\(\frac{-1}{2}\), ta có:
\(\left(\frac{-1}{2}\right)^2+4.\left(\frac{-1}{2}\right)+3=\frac{1}{4}+\left(-2\right)+3=\frac{5}{4}\)
b. Ta có:
\(x^2+4x+3=0\)
\(\Rightarrow x^2+x+3x+3=0\)
\(\Rightarrow\left(x^2+x\right)+\left(3x+3\right)=0\)
\(\Rightarrow x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+1=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=-3\end{cases}}}\)
Vậy \(x=-1;x=-3\)
tinh gia tri bieu thuc ;a+a+a+a+a+b+b+b+b+b.biet a + b=105
gia tri cua bieu thuc tren la...
=(a+b)+(a+b)+(a+b)+(a+b)+(a+b)
=(a+b)*5
=105*5
=525
bai1 :P=x+2/x+3-5/(x+3)(x-2)+1/2-x
a,Tim dkxd cua P
b,Rut gon bieu thuc P
c,tim x de P=-3/4
d,tim gia tri nguyen cua x de P cung co gia tri nguyen
e,tinh gia tri bieu thuc P khi x^2-9=0
cho a/2=b/3=c/4. tinh gia tri bieu thuc a^2+b^2+2c^2/a^2-4b^2+c^2
Giải:
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\)
\(\Rightarrow a=2k,b=3k,c=4k\)
Ta có: \(\frac{a^2+b^2+2c^2}{a^2-4b^2+c^2}\)
\(=\frac{\left(2k\right)^2+\left(3k\right)^2+2\left(4k\right)^2}{\left(2k\right)^2-4\left(3k\right)^2+\left(4k\right)^2}\)
\(=\frac{2^2.k^2+3^2.k^2+2.4^2.k^2}{2^2.k^2-4.3^2.k^2+4^2.k^2}\)
\(=\frac{4.k^2+9.k^2+32.k^2}{4.k^2-36.k^2+16.k^2}\)
\(=\frac{k^2.\left(4+9+32\right)}{k^2.\left(4-36+16\right)}\)
\(=\frac{45}{-16}\)
\(A=\frac{a^2+b^2+2c^2}{a^2-4b^2+c^2}\)
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\)
Suy ra \(A=\frac{\left(2k\right)^2+\left(3k\right)^2+2\left(4k\right)^2}{\left(2k\right)^2-4\left(3k\right)^2+\left(4k\right)^2}=\frac{4k^2+9k^2+2\cdot16k^2}{4k^2-4\cdot9k^2+16k^2}\)
\(=\frac{k^2\left(4+9+32\right)}{k^2\left(4-36+16\right)}=\frac{45}{-16}=-\frac{45}{16}\)
cho bieu thuc A=2,45*(a+64,36)-45
a tinh gia tri bieu thuc A khi a=35,64
b tim gia tri cua A de bieu thuc A =126,5
b. Neu A =126.5 thì 126.5=2.45 x (a+64.36) -45
= 2.45 x (a+64.36) 126.5 + 45
= 2.45 x (a+64.36) = 171.5
= a+64.36 = 171.5 : 24.5
a+64.36 = 70
a = 70 - 64.36
a = 5.64
a,thay a=35,64vào bthuc đã cho ta đc:
A= 2,45x(35,64 +64,36)-45=2,45x100-45=200
b,để A=126,5 thì 2,45x(a+64,36)-45=126,5
=>2,45a+157,682-45=126,5
=>2,45a=13,818
=>a=5,64
cistian ronaldo