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\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+301\)
\(=301-4\cdot75=301-300=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{1}{358}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right).2.\frac{1}{297}-7.\frac{1}{358}-\frac{3}{297}.\frac{1}{358}\)
\(=\frac{1}{358}.\left(7+\frac{1}{297}-7-\frac{3}{297}\right)-\left(4-\frac{1}{358}\right).\frac{2}{297}\)
\(=\frac{1}{358}.\left(-\frac{2}{297}\right)-\frac{2}{297}.\left(4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(\frac{1}{358}+4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(-4\right)\)
\(=\frac{8}{297}\)
Vậy giá trị biểu thức A là \(\frac{8}{297}\)
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Bài 1:
\(A=23^2+46\cdot37+37^2=23^2+2\cdot23\cdot37+37^2=\left(23+37\right)^2=60^2=3600\)
\(B=27^2-44\cdot27+22^2=27^2-2\cdot27\cdot22+22^2=\left(27-22\right)^2=5^2=25\)
Bài 2:
\(A=x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1\)
Vì: \(\left(x-2\right)^2\ge0\) với mọi x
=> \(\left(x-2\right)^2+1\ge1\)
Vậy GTNN của A là 1 khi x=2
\(A=23^2+2.23.37+37^2=\left(23+37\right)^2=60^2=3600\)
\(B=27^2-2.27.22+22^2=\left(27-22\right)^2=5^2=25\)
\(A=x^2-4x+5=\left(x-2\right)^2+1\ge1\)
=> A min=1 khi x=2
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\(Q=\dfrac{23-10x}{x^2+2}=\dfrac{46-20x}{2\left(x^2+2\right)}=\dfrac{25\left(x^2+2\right)-25x^2-20x-4}{2\left(x^2+2\right)}\)
\(=\dfrac{25}{2}-\dfrac{\left(5x+2\right)^2}{2\left(x^2+2\right)}\le\dfrac{25}{2}\)
\(Q_{max}=\dfrac{25}{2}\) khi \(5x+2=0\Rightarrow x=-\dfrac{2}{5}\)
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a) Rút gọn thu được B = 4 x ( 2 + x ) ( 2 − x ) ( 2 + x ) : x − 3 x ( 2 − x ) = 4 x 2 x − 3 với x ≠ ± 2 ; x ≠ 0 ; x ≠ 3
b) 4 x 2 x − 3 < 0 ⇔ x − 3 < 0 ⇔ x < 3 ;
Kết hợp điều kiện được 0 < x < 3; x ≠ ± 2.
299 - 298 - 297 -.........- 2 - 1
= -(299 + 298 + 297 +....+2+1)
Đặt A = 299 + 298 + 297 +....+2+1
2A = 2100 + 299 + 298 +...+ 22 + 2
A = 2A - A = 2100 - 1
=> 299 - 298 - 297 -.........- 2 - 1
= -(2100 - 1)
= -2100 + 1