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\(\left(x-3\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)
\(\Rightarrow x\in\left\{3;12\right\}\)
\(\left(x^2-81\right)\left(x^2+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)
\(\Rightarrow x=9\)
\(\left(x-4\right)\left(x+2\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu
\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)
\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Vậy \(x\in\left\{-1;0;1;2;3\right\}\)
Bài 1:
a, x = 0
b, x = 2
c, x = 1
Bài 2:
Nếu n=0 thì nên bới giá trị -n2
\(5x+2x=6^2-5^0\Leftrightarrow5x+2x=6^2-1\)
\(\Leftrightarrow5x-2x=36-1\Leftrightarrow5x-2x=35\Leftrightarrow x\left(5+2\right)=35\)
\(\Leftrightarrow7x=35\Leftrightarrow x=35:7\Leftrightarrow x=5\)
\(5x+x=150:2+3\Leftrightarrow5x+x=75+3\)
\(\Leftrightarrow5x+x=78\Leftrightarrow x\left(5+1\right)=78\Leftrightarrow x6=78\)
\(\Leftrightarrow x=78:6\Leftrightarrow x=13\)
a) 5x + 2x = 62 - 50
7x = 35 => x = 5
b) 5x + x = 150 : 2 + 3
6x = 78 => x = 13
c) 6x + x = 511 : 59 + 31
7x = 28 => x = 4
d) 5x + 3x = 36 : 33 x 4 + 12
8x = 120 => x = 15
e) 4x + 2x = 68 - 219 : 216
6x = 60 => x = 10
f) 5x + x = 39 - 311 : 39
6x = 30 => x = 5
g) 7x - x = 521 : 519 + 3 x 22 - 70
6x = 36 => x = 6
h) 7x - 2x = 617 : 615 + 44 : 11
5x = 40 => x = 8
Bài làm
m) (x + 2).(3 - x) = 0;
=> x + 2 = 0 hoặc 3 - x = 0
=> x = -2 hoặc x = 3
Vậy x = -2 hoặc x = 3
d) 511.712 + 511.711
= 511 . ( 712 + 711 )
= 511 . [ 711 . ( 7 + 1 ) ]
= 511 . 711 . 8
= ( 5 . 7 )11 . 8
= 3511 . 8
512.712 + 9.511.711
= 511 ( 5 . 712 + 9 . 1 . 711 )
= 511 [ 711 ( 5 . 7 + 9 . 1 . 1 ) ]
= 511 ( 711 . 44 )
= 511 . 711 . 44
= 3511 . 44
m. \(\left(x+2\right)\left(3-x\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\3-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
d. \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.\left(7^{12}+7^{11}\right)}{5^{11}.\left(5.7^{12}+9.7^{11}\right)}=\frac{7^{12}+7^{11}}{5.7^{12}+9.7^{11}}=\frac{1}{5.9}=\frac{1}{45}\)
q. \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10+11=11\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10=0\)
\(\Rightarrow\left[\left(x-3\right)+\left(x-2\right)+\left(x-1\right)\right]+(1+2+3+...+10)=0\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+55=0\)
\(\Rightarrow x-3+x-2+x-1=-55\)
\(\Rightarrow3x-6=-55\)
\(\Rightarrow3x=-49\)
\(\Rightarrow x=-\frac{49}{3}\)
a) (2010 - x)(x + 2021) = 0
<=> \(\left[{}\begin{matrix}2010-x=0\\x+2021=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=2010\\x=-2021\end{matrix}\right.\)
b) (x - 2)^2 + 5(x - 2) = 0
<=> x^2 - 4x + 4 + 5x - 10 = 0
<=> x^2 + x - 6 = 0
<=> x^2 - 2x + 3x - 6 = 0
<=> x(x - 2) + 3(x - 2) = 0
<=> (x + 3)(x - 2) = 0
<=> \(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)