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\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)

=>\(\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

=>x-2010=0

=>x=2010

(x - 1)/2009 + (x - 2)/2008 = (x - 3)/2007 + (x - 4)/2006

(x - 1)/2009 - 1 + (x - 2)/2008 - 1 = (x - 3)/2007 - 1 + (x - 4)/2006 - 1

(x - 2010)/2009 + (x - 2010)/2008 = (x - 2010)/2007 + (x - 2010)/2006

(x - 2010)/2009 + (x - 2010)/2008 - (x - 2010)/2007 - (x - 2010)/2006 = 0

(x - 2010).(1/2009 + 1/2008 - 1/2007 - 1/2006) = 0

x - 2010 = 0

x = 2010

\(\left|x-3\right|+\left|2y-10\right|=0\)

Mà: \(\left\{{}\begin{matrix}\left|x-3\right|\ge0\forall x\\\left|2y-10\right|\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left|x-3\right|+\left|2y-10\right|=0\forall x,y\)

Mặt khác: \(\left|x-3\right|+\left|2y-10\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\2y-10=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\2y=10\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\end{matrix}\right.\)

Vậy: \(\left(x;y\right)=\left(3;5\right)\)

\(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)^2\)

\(=\left(2x\right)^2\)

\(=4x^2\)

2022/2023 . (9/13 - 7/11) + 2022/2023 . (17/13- 4/17)

= 2022/2023 . 190/43 + 2022/2023 . 237/221

= 2022/2023 . (190/43 + 237/221)

= 2022/2023 . 52181/9503

= 105509982/19224569

Sửa: \(\dfrac{2022}{2023}\cdot\left(\dfrac{9}{13}-\dfrac{7}{11}\right)+\dfrac{2022}{2023}\cdot\left(\dfrac{17}{13}-\dfrac{4}{11}\right)\)

\(=\dfrac{2022}{2023}\cdot\left(\dfrac{9}{13}-\dfrac{7}{11}+\dfrac{17}{13}-\dfrac{4}{11}\right)\)

\(=\dfrac{2022}{2023}\cdot\left(2-1\right)\)

\(=\dfrac{2022}{2023}\cdot1\)

\(=\dfrac{2022}{2023}\)

a: Xét ΔAHB vuông tại H và ΔAHC vuông tại H có

AB=AC

AH chung

Do đó: ΔAHB=ΔAHC

b: ΔAHB=ΔAHC

=>\(\widehat{BAH}=\widehat{CAH}\)

=>AH là phân giác của \(\widehat{BAC}\)

c: ΔAHB=ΔAHC

=>HB=HC

=>H là trung điểm của BC

Điểm trung bình môn Toán học kì 1 của bạn Cường là:

\(\dfrac{7+8+6+10+2\cdot7+2\cdot6+2\cdot5+2\cdot9+8\cdot3}{1\cdot4+2\cdot4+3}\)

\(=\dfrac{15+16+2\cdot\left(7+6+5+9\right)+24}{7+8}\)

\(=\dfrac{55+2\cdot27}{15}\simeq7,3\)

c) \(C=\dfrac{12}{21-\left|3x+4\right|}\left(dkxd:x\ne\dfrac{17}{3};x\ne-\dfrac{25}{3}\right)\)

Ta thấy: \(\left|3x+4\right|\ge0\)

\(\Rightarrow-\left|3x+4\right|\le0\forall x\)

\(\Rightarrow21-\left|3x+4\right|\le21\forall x\)

\(\Rightarrow\dfrac{1}{21-\left|3x+4\right|}\ge\dfrac{1}{21}\forall x\)

\(\Rightarrow C=\dfrac{12}{21-\left|3x+4\right|}\ge\dfrac{12}{21}=\dfrac{4}{7}\forall x\)

Dấu \("="\) xảy ra khi: \(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)

Vậy \(Min_C=\dfrac{4}{7}\) khi \(x=-\dfrac{4}{3}\).

c) C = 12 21 − | 3 x + 4 | ( d k x d : x ≠ 17 3 ; x ≠ − 25 3 ) Ta thấy: | 3 x + 4 | ≥ 0 ⇒ − | 3 x + 4 | ≤ 0 ∀ x ⇒ 21 − | 3 x + 4 | ≤ 21 ∀ x ⇒ 1 21 − | 3 x + 4 | ≥ 1 21 ∀ x ⇒ C = 12 21 − | 3 x + 4 | ≥ 12 21 = 4 7 ∀ x Dấu "=" xảy ra khi: 3 x + 4 = 0 ⇔ x = − 4 3 ( t m ) Vậy M i n C = 4 7 khi x = − 4 3 .

\(\left|6x-2\right|>=0\)

=>\(\left|6x-2\right|-15>=-15\)

=>\(B=\dfrac{5}{\left|6x-2\right|-15}< =\dfrac{5}{-15}=-\dfrac{1}{3}\)

Dấu = xảy ra khi 6x-2=0

=>\(x=\dfrac{1}{3}\)

Ta thấy:

\(\left|2x-5\right|\ge0\forall x\)

\(\left|3y+1\right|\ge0\forall y\)

\(\Rightarrow\left|2x-5\right|+\left|3y+1\right|\ge0\forall x;y\)

\(\Rightarrow A=\left|2x-5\right|+\left|3y+1\right|-17\ge-17\forall x;y\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(Min_A=-17\) khi \(x=\dfrac{5}{2};y=-\dfrac{1}{3}\).

\(\left|2x-5\right|>=0\forall x;\left|3y+1\right|>=0\forall y\)

=>\(\left|2x-5\right|+\left|3y+1\right|>=0\forall x,y\)

=>\(A=\left|2x-5\right|+\left|3y+1\right|-17>=-17\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)