Biết l i m x → 0 3 x 2 + 2 - 2 - 2 x x = a 2 b ( a b tối giản). Giá trị của a + b bằng:
A. - 1 2
B. 3
C. 1 2
D. 2
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Bài 1:
a) \(\left(m+2\right).3-5=4\)
\(\Leftrightarrow3m+6-5=4\)
\(\Leftrightarrow3m+1=4\)
\(\Leftrightarrow3m=4-1\)
\(\Leftrightarrow3m=3\)
\(\Leftrightarrow m=1\)
Vậy: m = 1
b) \(\left(m-3\right).\left(-2\right)+8=-10\)
\(\Leftrightarrow-2m+6+8=-10\)
\(\Leftrightarrow-2m+14=-10\)
\(\Leftrightarrow-2m=-10-14\)
\(\Leftrightarrow-2m=-24\)
\(\Leftrightarrow m=12\)
Vậy: m = 12
Bài 2:
a) \(\left(x-2\right)^2=9\)
\(\Leftrightarrow\left(x-2\right)^2=3^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
b) \(\left(x+3\right)^2-0,16=0\)
\(\Leftrightarrow\left(x+3\right)^2=0,16\)
\(\Leftrightarrow\left(x+3\right)^2=\left(0,4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0,4\\x+3=-0,4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2,6\\x=-3,4\end{matrix}\right.\)
c) \(x^3=25x\)
\(\Leftrightarrow x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm5\end{matrix}\right.\)
Câu 3:
\(\Leftrightarrow\left(x-2\right)^{x+2010}\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-2\right)^{x+2010}\cdot\left(x-3\right)\left(x-1\right)=0\)
hay \(x\in\left\{1;2;3\right\}\)
Bài 2:
a: =>x+32=0
=>x=-32
b: =>x-1=0
=>x=1
c: =>45-x=0 hoặc x=0
=>x=0 hoặc x=45
d: =>x-12=0 hoặc x+27=0
=>x=12 hoặc x=-27
Bài 1:
a: =>(x-1-2)(x-1+2)=0
=>(x+1)(x-3)=0
=>x=3 hoặc x=-1
b: =>(x-3)(2x-x-3)=0
=>(x-3)(x-3)=0
=>x=3
c: =>x^3-1=5x+x^3-5-x^2
=>-x^2+5x-5=1
=>-x^2+5x-6=0
=>x^2-5x+6=0
=>x=2 hoặc x=3
bài 1
a, \(x^2+9y^2-6xy=\left(x-3y\right)^2\)
thay x = 19 , y = 3 vào biểu thức trên ta có
\(\left(19-3.3\right)^2=100\)
b, \(x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)
thay x = 12 và y = -4 vào biểu thức trên ta có
\(\left(12-2.\left(-4\right)\right)^3=8000\)
bài 4
a, \(x\left(4x^2-1\right)=0\)
=> \(x\left(2x-1\right)\left(2x+1\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\2x-1=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b, \(x^3-x^2-x+1=0\)
=> \(x^2\left(x-1\right)-\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c, \(2x^2-5x-7=0\)
=> \(2x^2-7x+2x-7=0\)
=> \(2x\left(x+1\right)-7\left(x+1\right)=0\)
=> \(\left(x+1\right)\left(2x-7\right)=0\)
=> \(\left[{}\begin{matrix}x+1=0\\2x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{2}\end{matrix}\right.\)
Bài 2: Rút gọn biểu thức:
a) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2+4xy+2y^2-x^2+y^2\)
\(=2y^2-2xy\)
b)\(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(2x+5\right)^2-3\left(1+4x\right)\left(1-4x\right)\)
\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)
\(=8x^2+40x+50-3+48x^2\)
\(=56x^2+40x+47\)
Bài 1:
\(\Leftrightarrow x^2+x-6-x^2+1=7\)
=>x-5=7
hay x=12
Phần a câu viết rỗ đề cho tớ nha !
b) \(\dfrac{8}{9}\)x( 2x2 - 3) = 0
*) \(\dfrac{8}{9}\)x = 0 --> x = 0
*) 2x2 - 3 = 0 --> 2x2 = 3 --> x2 = \(\dfrac{3}{2}\)--> x = \(\sqrt{\dfrac{3}{2}}\)
Vậy .....
Bài 1.
a, (x-2)-15=65
x-2=65+15
x-2=80
x=80+2
x=2
b, 115-2\(\times\)(x-3)=35
2\(\times\)(x-3)=115-35
2\(\times\)(x-3)=70
x-3=70:2
x-3=35
x=35+5
x=38
c, 35+2\(\times\)(x-3)=65
2\(\times\)(x-3)=65-35
2\(\times\)(x-3)=30
x-3=30:2
x-3=15
x=15+3
x=18
3\(\times\)(x-5)-16=11
3\(\times\)(x-5)=11+16
3\(\times\)(x-5)=27
x-5=27:3
x-5=9
x=9+5
x=14
Bài 2:
a, \(2^x-1=31\)
\(2^x=31-1\)
\(2^x=30\)
\(\Rightarrow\)Không có x thoả mãn điều kiện \(2^x=30\)
b, \(x^3-1=26\)
\(x^3=26+1\)
\(x^3=27\)
\(\Rightarrow x=3\) vì \(3^3=27\)
c, \(6^x-1+1=37\)
\(6^x-1=37-1\)
\(6^x-1=36\)
\(6^x=36+1\)
\(6^x=37\)
\(\Rightarrow\) Không có x thoả mãn điều kiện \(6^x=37\)
d, (x+2)\(^3\)-15\(^0\)=215
\(\left(x+2\right)^3-1=215\)
\(\left(x+2\right)^3=215+1\)
\(\left(x+2\right)^3=216\)
\(\left(x+2\right)^3=6^3\)
\(x+2=6\)
\(x=6-2\)
\(x=4\)
e, \(2\times\left(x-9\right)^2=2\)
\(\left(x-9\right)^2=2:2\)
\(\left(x-9\right)^2=1\)
\(\Rightarrow x-9=1\) vì \(1^2=1\)
x=1+9
x=10
g, \(3\times\left(x-5\right)^3=51\)
\(\left(x-5\right)^3=51:3\)
\(\left(x-5\right)^3=17\)
\(\Rightarrow\) Không có x thoả mãn điều kiện \(\left(x-5\right)^3=17\)
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Bài 1:
a)\(\left(x-2\right)-15=65\)
\(x-2=65+15\)
\(x-2=80\)
\(x=80+2\)
\(x=82\)
b)\(115-2\left(x-3\right)=35\)
\(2\left(x-3\right)=115-35\)
\(2\left(x-3\right)=80\)
\(x-3=80:2\)
\(x-3=40\)
\(x=40+3\)
c) \(35+2\left(x-3\right)=65\)
\(2\left(x-3\right)=65-35=30\)
\(x-3=30:2=15\)
\(x=15+3=18\)
d) \(3\left(x-5\right)-16=11\)
\(3\left(x-5\right)=11+16=27\)
\(x-5=27:3=9\)
\(x=9+5=14\)
Bài 2:
a) \(2^x-1=31\)
\(2^x=31+1=32\)
Vì \(2^5=32\Rightarrow x=5\)
b) \(x^3-1=26\)
\(x^3=26+1=27\)
Vì \(3^3=27\Rightarrow x=3\)
c)\(6^{x-1}+1=37\)
\(6^{x-1}=37-1=36\)
Vì \(6^6=36\Rightarrow x-1=6\Rightarrow x=6+1=7\)
d)\(\left(x+2\right)^3-15^0=215\)
\(\left(x+2\right)^3-1=215\)
\(\left(x+2\right)^3=215+1=216\)
Vì \(6^3=216\Rightarrow x+2=6\Rightarrow x=6-2=4\)
e)\(2\left(x-9\right)^2=2\)
\(\left(x-9\right)^2=2:2=1\)
Vì \(1^2=1\Rightarrow x-9=1\Rightarrow x=1+9=10\)
g) \(3\left(x-5\right)^3=51\)
\(\left(x-5\right)^3=51:3=17\)
Đáp án B