tinh Gia tri cua bieu thuc A=a^4-4a^3+a^2+6a+4/(a^2-2a+12) tai a= can cua 5 +1
Những câu hỏi liên quan
cho bieu thuc a=-1/3+1/3^2-1/3^3+1/3^4-1/3^5+...+1/3^100 tinh gia tri cua bieu thuc b=4/a/+1/3^100
Cho x+y=2, tinh gia tri cua bieu thuc:
M=3(x^2+y^2)-(x^3+y^3)+1
Bai 2:Cho a+b=5,tinh gia tri bieu thuc:
M=3a^2-2a+3b^2-2b+6ab+100
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: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
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Tinh gia tri cua bieu thuc :
a) A= 5a^3 b^4 voi a = -1, b = 1
b) B = 9a^5 b^2 voi a = -1, b = 2
Giup minh voi minh dang can cuc gap.
cho bieu thuc P=\(\dfrac{^{\left(a+3\right)^2}}{2a^26a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)
a.tim dieu kien xax dinh cua P
b.rut gon bieu thuc P
c.voi gia tri nao cua a thi P=0;P=1
a) ĐKXĐ: \(a\ne0\) ; \(a\ne3\) ; \(a\ne-3\)
b) \(P=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\left(\dfrac{a^2-9}{a^2-9}-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{\left(a^2-9\right)-\left(6a-18\right)}{a^2-9}\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{a^2-9-6a+18}{a^2-9}\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{a^2-6a+9}{a^2-9}\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{\left(a-3\right)^2}{\left(a-3\right)\left(a+3\right)}\)
\(\Leftrightarrow P=\dfrac{a+3}{2a}.\dfrac{a-3}{a+3}\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)\left(a-3\right)}{2a\left(a+3\right)}\)
\(\Leftrightarrow P=\dfrac{a-3}{2a}\)
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( ko biết đúng hay ko)
c) \(P=\dfrac{a-3}{2a}=0\)
\(\Leftrightarrow a-3=0\)
\(\Leftrightarrow a=3\left(loai\right)\) ( không thỏa mãn điều kiện )
\(P=\dfrac{a-3}{2a}=1\)
\(\Leftrightarrow a-3=2a\)
\(\Leftrightarrow a-3-2a=0\)
\(\Leftrightarrow-a-3=0\)
\(\Leftrightarrow-a=3\)
\(\Leftrightarrow a=-3\left(loai\right)\) ( không thỏa mãn điều kiện )
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Tinh
Xy^3+4xy^3-3xy^3
(-4/5ab^2c)×(-20a^4b^3c)
Bai 2.tinh gia tri cua bieu thuc a=14x^2+5xy-2010y^2 tai x=-1;y=-2
xy3+4xy3-3xy3
=5xy3-3xy3 = 2xy3
tươg tự
Bài 2 : Thay zô có j kó đâu ==
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cho bieu thuc A=[x+2/x^2-x+x-2/x^2+x].x^2-1/x^2+2
a) tim dieu kien cua x de gia tri cua bieu thuc A duoc xac dinh
b) tinh gia tri cua bieu thuc A voi x = -200
a) \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2-x}\)
\(A=\left[\dfrac{x+2}{x\left(x-1\right)}+\dfrac{x-2}{x\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{x^2+2x+x+2+x^2-2x-x+2}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2x^2+4}{x\left(x^2-1\right)}.\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2\left(x^2+2\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x^2+2\right)}=\dfrac{2}{x}\)
b) Thay \(x=-200\) vào biểu thức \(A=\dfrac{2}{x}\) ta được :
\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)
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tinh gia tri cua cac bieu thuc
A=3a-2b\a-3b voi a\b=10\3
B=a-8\b-5 - 4a-b\3a+a voi a-b = 3 va b khac 5 b khac -4
Cho bieu thuc A=\(\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\div\dfrac{1}{\sqrt{x}-1}\)
a/ Tim dieu kien cua x de bieu thuc A co gia tri xac dinh
b/ Rut gon A
c/ Tinh gia tri cua A khi x = \(4-2\sqrt{3}\)
d/ Tim gia tri nho nhat cua A
a. ĐKXĐ : x>1.
b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)
c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:
\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)
Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).
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