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a) \(\frac{19}{27}+\frac{9}{7}+\frac{6}{27}+\frac{5}{7}+\frac{2}{27}\)
\(=\left(\frac{19}{27}+\frac{6}{27}+\frac{2}{27}\right)+\left(\frac{9}{7}+\frac{5}{7}\right)\)
\(=1+2=3\)
b) \(\frac{2}{7}+\frac{27}{48}+\frac{7}{16}+\frac{45}{63}\)
\(=\frac{2}{7}+\frac{9}{16}+\frac{7}{16}+\frac{5}{7}\)
\(=\left(\frac{2}{7}+\frac{5}{7}\right)+\left(\frac{9}{16}+\frac{7}{16}\right)\)
\(=1+1=2\)
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a) \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2.\left(-6\right)=13\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3.\left(-6\right).1=19\)
\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)=13.19-\left(-6\right)^2.1=211\)
b) \(x^2+y^2=\left(x-y\right)^2+2xy=1^1+2.6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+3.6.1=19\)
\(x^5-y^5=\left(x^2+y^2\right)\left(x^3-y^3\right)+x^2y^2\left(x-y\right)=13.19+6^2.1=283\)
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A = ( 15 : x + 15 * x ) + 2009 * y
= 30 * x + 2009 * y
Thay x = 1 và y = 0 vào A ta có :
A = 30 x 1 + 2009 x 0
= 30 + 2009
= 2039
B = y : ( 119 * x + 4512 ) + ( 756 : x - y )
Thay x = 1 và y = 0 vào B ta có :
B = 0 : ( 119 x 1 + 4512 ) + ( 756 : 1 - 0 )
= 0 + 756
= 756
C = ( 10 + y ) : ( 10 - y ) - x + 10 * x - 10 : x + y
Thay x = 1 và y = 0 vào C ta có :
C = ( 10 + 0 ) : ( 10 - 0 ) - 1 + 10 x 1 - 10 : 1 + 0
= 10 : 10 - 1 + 10 - 10
= 1 - 1 + 10 - 10
= 0
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1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
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\(F=-x^2-2y^2+2xy-y+1\)
\(-F=x^2+2y^2-2xy+y-1\)
\(-F=\left(x^2-2xy+y^2\right)+\left(y^2+y+\frac{1}{4}\right)-\frac{5}{4}\)
\(-F=\left(x-y\right)^2+\left(y+\frac{1}{2}\right)^2-\frac{5}{4}\)
Mà \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(y+\frac{1}{2}\right)^2\ge0\forall y\)
\(\Rightarrow-F\ge-\frac{5}{4}\)
\(\Leftrightarrow F\le\frac{5}{4}\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x-y=0\\y+\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{1}{2}\end{cases}}\)
Vậy \(F_{Max}=\frac{5}{4}\Leftrightarrow x=y=-\frac{1}{2}\)
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Lời giải:
a. $99^3+1+3(99^2+99)=99^3+3.99^2.1+3.99.1^2+1^3=(99+1)^3=100^3=1000000$
b. $11^3-1-3(11^2-11)=11^3-3.11^2.1+3.11.1^2-1^3=(11-1)^3=10^3=1000$
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Đây nè bạn. Mk chỉ mới nghĩ ra cách này thôi à!!! Bạn nào có cách nào thì bảo mk với nhé!!!
\(B=4\left(x+y\right)\left(x^2+y^2-xy\right)-6\left[\left(x+y\right)^2-2xy\right]\)
\(=4\left[\left(x+y\right)^2-3xy\right]-6\left(1-2xy\right)\)
\(=4-12xy-6+12xy\)
\(=-2\)
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a/
\(x-y=\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}=\frac{1}{bd}.\) (1)
\(y-z=\frac{c}{d}-\frac{e}{h}=\frac{ch-de}{dh}=\frac{1}{dh}\)(2)
+ Nếu d>0 => (1)>0 và (2)>0 => x>y; y>x => x>y>z
+ Nếu d<0 => (1)<0 và (2)<0 => x<y; y<z => x<y<z
b/
\(m-y=\frac{a+e}{b+h}-\frac{c}{d}=\frac{ad+de-cb-ch}{d\left(b+h\right)}=\frac{\left(ad-cb\right)-\left(ch-de\right)}{d\left(b+h\right)}=\frac{1-1}{d\left(b+h\right)}=0\)
=> m=y
+
cảm ơn bn nha Nguyễn Ngoc Anh Minh mk k cho bn r đó kb vs mk nha
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bằng 1