Tìm a,b \(\in\) Z thoả
(a-1)2 \(\times\) (a2+9)=4\(\times\)b2 +20\(\times\)b +25
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a, \(\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=0+5=5\\x=0-2=-2\end{matrix}\right.\)
Vậy x = 5 hoặc x = -2
b, \(26\left(2x+4\right)\left(x-1\right)=0\Leftrightarrow\left(2x+4\right)\left(x-1\right)=0\)
\(\Rightarrow\left[\begin{matrix}2x+4=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}2x=-4\\x=1\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy x = -2 hoặc x = 1
c, \(\left(x^2-9\right)\left(x^2-25\right)< 0\)
\(\Rightarrow\) x2 - 9 và x2 - 25 trái dấu
Mà : \(x^2-9>x^2-25\)
\(\Rightarrow\left\{\begin{matrix}x^2-9>0\\x^2-25< 0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x^2>9\\x^2< 25\end{matrix}\right.\)\(\Rightarrow9< x^2< 25\)
Mà : \(x\in Z\) => x2 là số chính phương
\(x^2=16\Rightarrow x^2=\left(\pm4\right)^2\Rightarrow x=\pm4\)
Vậy \(x=\pm4\)
\(\left(x-5\right)\left(x+2\right)=0\)
=> x - 5 = 0 và x + 2 = 0
=> x = -5 và x = -2
\(A\cdot\left(1-\dfrac{1}{4}\right)\cdot\left(1-\dfrac{1}{9}\right)\cdot\left(1-\dfrac{1}{16}\right)\left(1-\dfrac{1}{25}\right)=1\dfrac{3}{5}\)
=>\(A\cdot\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{5}\right)=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{5}\cdot\dfrac{6}{2}=\dfrac{8}{5}\)
=>\(A\cdot3=8\)
=>A=8/3
a: th1: a>0
=>-17a<27a
th2: a=0
=>-17a=27a
th3: a<0
=>-17a>27a
b:
TH1: a>6
36>31
nên 36(a-6)>31(a-6)
TH2: a<6
36>31
nên 36(a-6)<31(a-6)
a, \(-29-9.\left(2x-1\right)^2=-110\)
\(\Rightarrow\left(2x-1\right)^2=\left[-29-\left(-110\right)\right]:9\)
\(\Rightarrow\left(2x-1\right)^2=9=3^2\)
\(\Rightarrow2x-1=9\)
\(\Rightarrow2x=9+1\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=10:2\)
\(\Rightarrow x=5\)
b, tương tự nha bạn
a) 13 x A + 5 x B + 13 x B + 5 x A
= ( 13 x A + 13 x B ) + ( 5 x B + 5 x A )
= 13 x ( A + B ) + 5 x ( B + A ) mà A + B = 12
=> 13 x ( A + B ) + 5 x ( B + A )
= 13 x 12 + 5 x 12
= 12 x ( 13 + 5 )
= 12 x 18
= 216
b) 4 x A + 25 x B + 16 x A - 5 x B
= ( 4 x A + 16 x A ) + ( 25 x B - 5 x B )
= A x ( 4 + 16 ) + ( 25 - 5 ) x B
= A x 20 + 20 x B
= 20 x ( A + B ) mà A + B = 12
=> 20 x ( A + B )
= 20 x 12
= 240
Trả lời :
a) 13 x A + 5 x B + 13 x B + 5 x A
= 13 x (A + B)+ 5 x (A + B)
= (13 x 12) + (5 x 12)
= 156 + 60
= 216
b) 4 x A + 25 x B + 16 x A - 5 x B
= 25 + 4 x (A + B) + 16 - 5 x (A + B)
= 29 x 12 + 11 x 12
= 348 + 132
= 480
T i c k nha !!