Tinh gia tri bieu thuc:a.3^2^3^0 b.5^2^0^3
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cho 2 bieu thuc:
A=(\(\sqrt{20}\) -\(\sqrt{45}\) +3\(\sqrt{5}\) ).\(\sqrt{5}\) va B=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}\) +\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) (Dieu kien: x>0, x khac 1
a) Rut gon bieu thuc A va B
b)Tim cac gia tri cua x de gia tri cua bieu thuc A bang 2lan gia tri B
a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
b: A=2B
=>\(10=4\sqrt{x}-2\)
=>\(4\sqrt{x}=12\)
=>x=9(nhận)
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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 \(x=-1;x=-3\)
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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
Tim cac gia tri a nguyen thoa man dang thuc:a^3+2a^2-5a-6=0
\(a^3+2a^2-5a-6=0\)
\(\Leftrightarrow a^3+a^2+a^2+a-6a-6=0\)
\(\Leftrightarrow a^2\left(a+1\right)+a\left(a+1\right)-6\left(a+1\right)=0\)
\(\Leftrightarrow\left(a+1\right)\left(a^2+a-6\right)=0\)
\(\Leftrightarrow\left(a+1\right)\left(a^2+3a-2a-6\right)=0\)
\(\Leftrightarrow\left(a+1\right)\left(a\left(a+3\right)-2\left(a+3\right)\right)=0\)
\(\Leftrightarrow\left(a+1\right)\left(a-2\right)\left(a+3\right)=0\)
\(\Leftrightarrow a=-1,a=2,a=-3\)
Vậy S={-3;-1;2}
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cho 2 bieu thuc:A=101 x50 ;B= 50x49+53x50
khong thuc hien tinh hay su dung tinh chat cua phep tinh de so sanh gia tri so cua A va B
Tinh gia tri bieu thuc sau:a.(2/3)^3^0-(5/3-3/4)^2. b.(53/4-59/27-65/6).230.1/25+187/4):(10/7+10/3):(37/7-100/7)
tinh gia tri bieu thuc sau: x^3+xy^2-x^2y-y^3+3 biet x-y=0
a, biet x+y=0
tinh gia tri bieu thuc : M=\(x^4-xy^3+x^3y-y^4-1\)
b, biet xyz=2 va x+y+z=0
tinh gia tri bieu thuc : M= \(\left(x+y\right)\left(y+2\right)\left(x+2\right)\)
a/ \(M=x^4-xy^3+x^3y-y^4-1\)
\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)
Mà \(x+y=0\)
\(\Leftrightarrow M=x^3.0-y^3.0-1\)
\(\Leftrightarrow M=-1\)
Vậy ...
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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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