Tinh' gia' tri. bieu thuc' ( x2 - 5 ) ( x + 3 ) + ( x + 4 ) ( x - x2 ) trong moi truong` hop sau.
a)x = 0
b)x = 15
c)x=-15
d)x=0.15
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Bài 2:
a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: Để A>0 thì x-3>0
hay x>3
Ta có: \(\Delta'=32>0\)
\(\Rightarrow\) Phương trình có 2 nghiệm phân biệt
Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=12\\x_1x_2=4\end{matrix}\right.\)
Mặt khác: \(T=\dfrac{x_1^2+x^2_2}{\sqrt{x_1}+\sqrt{x_2}}\)
\(\Rightarrow T^2=\dfrac{x_1^4+x^4_2+2x_1^2x_2^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(x_1^2+x_1^2\right)^2}{x_1+x_2+2\sqrt{x_1x_2}}\) \(=\dfrac{\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(12^2-2\cdot4\right)^2}{12+2\sqrt{4}}=1156\)
Mà ta thấy \(T>0\) \(\Rightarrow T=\sqrt{1156}=34\)
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 \(x=-1;x=-3\)
Theo bài ra, ta có: \(x^2-y=y^2-x\Leftrightarrow x^2-y^2=-x+y\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=-\left(x-y\right)\)
\(\Leftrightarrow\left(x+y\right)=-1\)
Ta lại có: \(A=x^2+2xy+y^2-3x-3y=\left(x+y\right)^2-3\left(x+y\right)\)
Thay x+y=-1 vào biểu thức A, ta được: \(A=\left(-1\right)^2-3.\left(-1\right)=1+3=4\)
Vậy A=4
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: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
A, 54 X 113 + 54 X 113 + 113
= 54 X 113 + 54 X 113 + 113x1
= (54+54+1)x113
= 109 x113
= 12317
A) 1000000 - 107492 : 154 x 28 + 1000
= 1000000 - 19544 + 1000
= 981456
B) 8/15 + 4/7 + 14/30 + 12/21 + 5
= 15/7 + 5
= 50/7
C) 9/13 : 6/11 - 5/12 : 6/11
= 9/13 x 11/6 - 5/12 x 11/6
= (9/13 - 5/12) x 11/6
= 43/156 x 11/6
= 473/936
Bài 2
1) X x2 + X x 1/5 = 8/5
X x (2 + 1/5 ) = 8/5
X x 11/5 = 8/5
X = 8/5 x 5/11
X = 8/11
2) X : 3/8 = 4/3 : 3/6
X : 3/8 = 8/3
X = 8/3 x 3/8
X = 1
a) 1 000 000 - 107 492 : 154 x 28 + 1000
= 1 000 000 - 698 x 28 + 1000
= 1 000 000 - 19544 + 1000
= 980 456 + 1000
= 981 456
b) 8/15 + 4/7 + 14/30 + 12/21 + 5
= ( 8/15 + 14/30 ) + ( 4/7 + 12/21 ) + 5
= ( 16/30 + 14/30 ) + ( 12/21 + 12/21 ) + 5
= 30/30 + 24/21 + 5
= 1 + 24/21 + 5
= ( 1 + 5 ) + 24/21
= 6 + 24/21
= 150/21
= 50/7