Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) \(x\left(5-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right..\)
Vậy .....
b ) \(\left(3-x\right)\left(x^2+1\right)=0\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy ........
c ) giống câu b
d ) \(\left(2x-1\right)^3=8\Leftrightarrow2x-1=2\Leftrightarrow\Leftrightarrow x=\dfrac{3}{2}\)
e ) \(\left(x+3\right)^2=16\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.
Bài 1:
a) \(2^8.2.4=2^9.2^2=2^{11}\)
b) \(8^5:64=8^5:8^2=8^3\)
c) \(3^7:9=3^7:3^2=3^5\)
d) \(9^{17}.81=9^{17}.9^2=9^{19}\)
e) \(x^6.x.x^2=x^9\)
Bài 2:
a) \(2^x-15=17\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5
b) \(2.3^x=162\)
\(3^x=162:2\)
\(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy x = 4
c) \(5.x.5^2=10\)
\(\Rightarrow x.5^3=10\)
\(\Rightarrow x.125=10\)
\(\Rightarrow x=10:125\)
\(\Rightarrow x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\)
d) \(5.x^2-1=124\)
\(\Rightarrow5.x^2=125\)
\(\Rightarrow x^2=125:5\)
\(\Rightarrow x^2=5^2\)
\(\Rightarrow x=\pm5\)
Vậy \(x=\pm5\)
Câu 1:
a)28.2.4=28.2.22=211
b)85:64=85:82=83
c)37:9=37:32=35
d)917.81=917.92=919
e)x6.x.x2=x9
Bài 1:
a: (x-1)(x-3)>=0
=>x-3>=0 hoặc x-1<=0
=>x>=3 hoặc x<=1
b: (x-5)(x-7)<0
=>x-5>0 và x-7<0
=>5<x<7
c: (x2-1)(x2-4)<0
=>1<x2<4
mà x là số nguyên
nên \(x\in\varnothing\)
\(\left(x-2\right)\left(x-4\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2< 0\\x-4>0\end{matrix}\right.=>4< x< 2\left(1\right)\\\left\{{}\begin{matrix}x-2>0\\x-4< 0\end{matrix}\right.=>2< x< 4\left(2\right)}\end{matrix}\right.\)(1 ) vô lý=> loại
=> (x-2).(x-4)<0 <=> 2<x<4
b. ta có\(x^2+1>0\forall x\)
=>(x2 -1).(x2+1)<0 <=> (x2 -1)<0 <=> x2<1
<=> -1<x<1
câu c bạn làm tương tự
nên các bn giúp mk nha,còn nhìu ý nữa cơ 
mai hoc rùi
Bài 1: Tính:
a) 27 : 22 + 54 : 53. 24 - 3. 25
= 25 + 5 . 24 - 3 . 25
= 32 + 5 . 16 - 3 . 32
= 32 + 80 - 96
= 112 - 96
= 16
b) ( 37 . 35) : 310+ 5 . 24 - 73 : 7
= 312 : 310 + 5 . 24 - 72
= 32 + 5 . 24 - 72
= 9 + 5 . 16 - 49
= 9 + 80 - 49
= 89 - 49
= 40
Bài 2: Tính hợp lí:
a) ( 62007 - 62006 ) : 62006
= 62007 : 62006 - 62006 : 62006
= 6 - 1
= 5
b) ( 112003 + 112002 ) : 112002
= 11 + 1
= 12
c) 320 : ( x3 - 24 ) + 24 = 32
320 : ( x3 - 24 ) = 32 - 24 = 8
x3 - 24 = 320 : 8
x3 - 24 = 40 + 24
x3 = 64
x3 = 43 = 4
d) 130 - ( 100 + x ) = 25
( 100 + x ) = 103 - 25
100 + x = 105 - 100
x = 5
Bn ơi đừng tự ti như vậy nha !!! Mỗi người đều có một khuyết điểm mà, tri thức luôn rộng lớn bao la. Hãy làm việc đó bằng cách bn tự làm những bài kia nha.
Chúc bn hc tốt môn toán :))
2)
a) \(\left(6^{2007}-6^{2006}\right):6^{2006}\)
\(=\left(6^{2006}.6-6^{2006}.1\right):6^{2006}\)
\(=\left[6^{2006}.\left(6-1\right)\right]:6^{2006}\)
\(=6^{2006}:6^{2006}.5\)
\(=5\)
b) \(\left(11^{2003}+11^{2002}\right):11^{2002}\)
\(=\left(11^{2002}.11+11^{2002}.1\right):11^{2002}\)
\(=\left[11^{2002}.\left(11+1\right)\right]:11^{2002}\)
\(=11^{2002}:11^{2002}.12\)
\(=12\)
c) \(130:\left(x^3-24\right)+24=32\)
\(\Leftrightarrow130:\left(x^3-24\right)=32-24\)
\(\Leftrightarrow130:\left(x^3-24\right)=8\)
\(\Leftrightarrow x^3-24=\dfrac{65}{4}\)
\(\Leftrightarrow x^3=\dfrac{65}{4}+24\)
\(\Leftrightarrow x^3=\dfrac{161}{4}\)
\(\Leftrightarrow x=\sqrt[3]{\dfrac{161}{4}}\)
Vậy \(x=\sqrt[3]{\dfrac{161}{4}}\)
d) \(130-\left(100+x\right)=25\)
\(\Leftrightarrow100+x=130-25\)
\(\Leftrightarrow100+x=105\)
\(\Leftrightarrow x=105-100\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
a. \(6^2:4.3+2.5^2\)
= \(36:12+2.25\)
= \(3+50\)
=\(53\)
b. \(2.\left(5.4^2-18\right)\)
= \(2.\left(5.16-18\right)\)
= \(2.\left(80-18\right)\)
= \(2.62\)
= \(124\)
c. \(80:\left\{\left[\left(11-2\right).2\right]+2\right\}\)
\(=80:\left\{\left[9.2\right]+2\right\}\)
\(=80:\left\{18+2\right\}\)
\(=80:20\)
\(=4\)
,bn nào giỏi toán giúp mk nha
a) \(3.5^2-16:2^3.2\)
\(=3.25-16:8.2\)
\(=75-2.2\)
\(=75-4\)
\(=71\)
b) \(168+\left\{\left[2\left(2^4+3^2\right)-256^0\right]:7^2\right\}\)
\(=168+\left\{\left[2\left(16+9\right)-256^0\right]:7^2\right\}\)
\(=168+\left[\left(2.25-256^0\right):7^2\right]\)
\(=168+\left[\left(50-1\right):7^2\right]\)
\(=168+\left(49:7^2\right)\)
\(=168+\left(49:49\right)\)
\(=168+1\)
\(=169\)
c) \(9^{20}:9^{18}-\left(4^2-7\right)^2+8.5^2+5600:\left(3^3+1^8\right)\)
\(=9^{20}:9^{18}-\left(16-7\right)^2+8.5^2+5600:\left(27+1\right)\)
\(=9^{20}:9^{18}-9^2+8.5^2+5600:28\)
\(=9^{20-18}-9^2+8.25+5600:28\)
\(=9^2-9^2+200+200\)
\(=81-81+200+200\)
\(=200+200\)
\(=400\)
Bài 1:
a) \(x\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
b) \(\left(3-x\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-0\\x^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\sqrt{-1}\end{matrix}\right.\)
Vì một số không âm mới có căn bậc hai \(\Rightarrow x=3\)
c) \(\left(x-1\right)^2=4\)
\(\Leftrightarrow\left(x-1\right)^2=2^2=\left(-2\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+1\\x=\left(-2\right)+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
d) \(\left(2x-1\right)^3=8\)
\(\Leftrightarrow\left(2x-1\right)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
\(\Leftrightarrow2x=2+1\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\) (không thỏa mãn vì \(x\in Z\))
Vậy \(x\in\varnothing\)
e) \(\left(x+3\right)^2=16\)
\(\Leftrightarrow\left(x+3\right)^2=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4-3\\x=\left(-4\right)-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)