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16 tháng 6 2018
a) Đặt \(A=4x-x^2-5\)
\(-A=x^2-4x+5\)
\(-A=\left(x^2-4x+4\right)+1\)
\(-A=\left(x-2\right)^2+1\)
Mà \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow-A\ge1\)
\(\Leftrightarrow A\le-1< 0\left(đpcm\right)\)
b) Đặt \(B=x^2-2x+5\)
\(B=\left(x^2-2x+1\right)+4\)
\(B=\left(x-1\right)^2+4\)
Mà \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow B\ge4>0\left(đpcm\right)\)
16 tháng 6 2018
a)4x-x2-5 = -(x2-4x+4)-1= -(x-2)^2 -1 < 0 với mọi x (đpcm)
b) x2 -2x+5= (x2-2x+1)+4=(x-1)^2 +4 >0 với mọi x (đpcm)
19 tháng 3 2020
a) Ta có: \(\left(x-1\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot3\cdot\left(x-2\right)=0\)
Vì 3≠0
nên \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: x∈{1;2}
b) Ta có: \(\left(2x+5\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\1-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-5\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{2}\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-5}{2};\frac{1}{3}\right\}\)
c) Ta có: \(\left(x+1\right)\left(2x-3\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=3\\3x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{3}{2};\frac{5}{3}\right\}\)
d) Ta có: \(6\left(x-2\right)\left(x-4\right)\left(1-7x\right)=0\)
Vì 6≠0
nên \(\left[{}\begin{matrix}x-2=0\\x-4=0\\1-7x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\\x=\frac{1}{7}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;4;\frac{1}{7}\right\}\)
e) Ta có: \(\left(x+1\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
Vậy: x∈{-1;-2}
f) Ta có: \(\left(3x-2\right)^2\cdot\left(x+1\right)\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-2\right)^2=0\\x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x=-1\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-1\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{2}{3};-1;2\right\}\)
g) Ta có: \(\left(5-x\right)^2\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(5-x\right)^2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5-x=0\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{1}{3}\right\}\)
h) Ta có: \(\left(14-2x\right)^2\cdot\left(3-x\right)\cdot\left(2x-4\right)=0\)
\(\Leftrightarrow4\left(7-x\right)^2\cdot\left(3-x\right)\cdot2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow8\cdot\left(7-x\right)^2\cdot\left(3-x\right)\cdot\left(x-2\right)=0\)
Vì 8≠0
nên \(\left[{}\begin{matrix}\left(7-x\right)^2=0\\3-x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7-x=0\\x=3\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\\x=2\end{matrix}\right.\)
Vậy: x∈{7;3;2}
i) Ta có: \(\left(5x-6\right)^2\cdot\left(x+2\right)\cdot\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(5x-6\right)^2=0\\x+2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x-6=0\\x=-2\\x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-2\\x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{5}\\x=-2\\x=-10\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{6}{5};-2;-10\right\}\)
j) Ta có: \(\left(3x-3\right)^3\cdot\left(x+4\right)=0\)
\(\Leftrightarrow27\cdot\left(x-1\right)^3\cdot\left(x+4\right)=0\)
Vì 27≠0
nên \(\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vậy: x∈{1;-4}
NN
15 tháng 10 2020
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
G
15 tháng 10 2020
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
LT
3
5 tháng 7 2019
a) \(9x^2-12x+4=0\)
\(\Leftrightarrow\left(3x\right)^2-3x.2.2+2^2=0\)
\(\Leftrightarrow\left(3x-2\right)^2=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy ...
b) \(\left(x-2\right)^2-25=0\)
\(\Leftrightarrow\left(x-2\right)^2=25\)
\(\Leftrightarrow\left(x-2\right)^2=5^2=\left(-5\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)
Vậy ...
5 tháng 7 2019
c) \(x^3+3x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)^3=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ...
PN
1
TM
22 tháng 7 2016
a)\(x\left(x+2\right)-3x-6=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>\(\left(x-3\right)\left(x+2\right)=0\)
=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
b)\(x^3+3x^2+3x-1-3x^2-3x=0\)
=>\(x^3-1=0\)
=>x3=1
=>x=1
TN
25 tháng 7 2016
Bài 1:
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1\(\ge\)0+1=1
Dấu = khi x+2=0 <=>x=-2
Vậy Amin=1 khi x=-2
b) B= ( x+3 ) ( x-11 ) + 2016
=x2-8x-33+2016
=x2-8x+16+1967
=(x-4)2+1967\(\ge\)0+1967=1967
Dấu = khi x-4=0 <=>x=4
Vậy Bmin=1967 <=>x=4
Bài 2:
a) D= 5 - 8x - x2
=-(x2+8x-5)
=21-x2+8x+16
=21-x2+4x+4x+16
=21-x(x+4)+4(x+4)
=21-(x+4)(x+4)
=21-(x+4)2\(\le\)0+21=21
Dấu = khi x+4=0 <=>x=-4
b)đề sai à
S
26 tháng 7 2016
ài 1:
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1$\ge$≥0+1=1
Dấu = khi x+2=0 <=>x=-2
Vậy Amin=1 khi x=-2
b) B= ( x+3 ) ( x-11 ) + 2016
=x2-8x-33+2016
=x2-8x+16+1967
=(x-4)2+1967$\ge$≥0+1967=1967
Dấu = khi x-4=0 <=>x=4
Vậy Bmin=1967 <=>x=4
Bài 2:
a) D= 5 - 8x - x2
=-(x2+8x-5)
=21-x2+8x+16
=21-x2+4x+4x+16
=21-x(x+4)+4(x+4)
=21-(x+4)(x+4)
=21-(x+4)2$\le$≤0+21=21
Dấu = khi x+4=0 <=>x=-4
b)đề sai à
30 tháng 7 2023
a: =>6x-3x^2-5=4-3x^2-2
=>6x-5=2
=>6x=7
=>x=7/6
b: =>20x+5-12x^2-3x=6x^2-10x+3x-5
=>-12x^2+17x+5-6x^2+7x+5=0
=>-18x^2+24x+10=0
=>x=5/3 hoặc x=-1/3
N
0
Đề bài là 3 . x2 hay là (3x)2 vậy?
\(3x^2+3x-5=0\)
Ta có: \(\Delta=3^2+4.3.5=69,\sqrt{\Delta}=\sqrt{69}\)
\(\Rightarrow\orbr{\begin{cases}x_1=\frac{-3+\sqrt{69}}{6}\\x_2=\frac{-3-\sqrt{69}}{6}\end{cases}}\)