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ĐKXĐ: \(-1\le x\le3\)
Đặt \(\sqrt{x+1}+\sqrt{3-x}=t\ge\sqrt{x+1+3-x}=2\)
\(\Rightarrow4+2\sqrt{-x^2+2x+3}=t^2\)
\(\Rightarrow\sqrt{-x^2+2x+3}=\dfrac{t^2-4}{2}\) (1)
Phương trình trở thành:
\(t-\dfrac{t^2-4}{2}=2\)
\(\Leftrightarrow2t-t^2=0\Rightarrow\left[{}\begin{matrix}t=0\left(loại\right)\\t=2\end{matrix}\right.\)
Thế vào (1):
\(\Rightarrow\sqrt{-x^2+2x+3}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
a) 5/17 * 8/-7+8/17*-7/3+-7/3*4/17
-40/119 + 12/17 × -7/3
-40/119 + -28/17 =-236/119
b) -10/13 + 5/17 - 3/13 + 12/17 - 11/20
(5/17+12/17)-(10/13+3/13)-11/20
-11/20
a) 5/17 * 8/-7+8/17*-7/3+-7/3*4/17
-40/119 + 12/17 × -7/3
-40/119 + -28/17 =-236/119
b) -10/13 + 5/17 - 3/13 + 12/17 - 11/20
(5/17+12/17)-(10/13+3/13)-11/20
-11/20
8 Her telephone number isn't known by me
9 The children will be brought home by my students
10 đúng r
11 We were given more information by her
12 All the workers of the plan were being instructed by the chief engineer
a) (a + b + c)2 = [(a + b) + c]2 = (a + b)2 + 2(a + b)c + c2
= a2+ 2ab + b2 + 2ac + 2bc + c2
= a2 + b2 + c2 + 2ab + 2bc + 2ac.
b) (a + b – c)2 = [(a + b) – c]2 = (a + b)2 - 2(a + b)c + c2
= a2 + 2ab + b2 - 2ac - 2bc + c2
= a2 + b2 + c2 + 2ab - 2bc - 2ac.
c) (a – b –c)2 = [(a – b) – c]2 = (a – b)2 – 2(a – b)c + c2
= a2 – 2ab + b2 – 2ac + 2bc + c2
= a2 + b2 + c2 – 2ab + 2bc – 2ac.
bài này phải không nếu đúng thì tích hộ mình
Bài 10: c)
\(C=8x^3-27y^3=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x-3y\right)\left[\left(2x-3y\right)^2+18xy\right]\)
\(=5.\left(5^2+18.4\right)=485\)
Bài 11:
\(\left(a+b+c\right)^3=\left(a+b\right)^3+c^3+3\left(a+b\right).c.\left(a+b+c\right)\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)\left(ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Bài 12:
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)^3-3c\left(a+b\right)\left(a+b+c\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ab-3bc-3ca\right]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)