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a) (2x2 - x) + 4x - 2 = 0
x(2x - 1) + 2(2x - 1) = 0
(2x - 1)(x + 2) = 0
2x - 1 = 0 hoặc x + 2 = 0
* 2x - 1 = 0
2x = 1
x = \(\frac{1}{2}\)
* x + 2 = 0
x = -2
Vậy x = -2; x = \(\frac{1}{2}\)
b) x2 - 6x + 8 = 0
x2 - 2x - 4x + 8 = 0
(x2 - 2x) + (-4x + 8) = 0
x(x - 2) - 4(x - 2) = 0
(x - 2)(x - 4) = 0
x - 2 = 0 hoặc x - 4 = 0
* x - 2 = 0
x = 2
* x - 4 = 0
x = 4
Vậy x = 2; x = 4
c) x4 - 8x2 - 9 = 0
x4 + x2 - 9x2 - 9 = 0
(x4 - 9x2) + (x2 - 9) = 0
x2(x2 - 9) + (x2 - 9) = 0
(x2 - 9)(x2 + 1) = 0
x2 - 9 = 0 (vì x2 + 1 > 0 với mọi x)
x2 = 9
x = 3 hoặc x = -3
Vậy x = 3; x = -3
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
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
a) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b) Ta có: \(-x^2+5x-6=0\)
\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left[\left(x-2\right)\left(x-3\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: x∈{2;3}
c) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
⇔(4x2-10x)-(2x-5)=0
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
d) Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)
e) Ta có: \(x^3+2x^2-x-2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1;-1\right\}\)
g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)
\(\Leftrightarrow-24x-8=0\)
\(\Leftrightarrow-8\left(3x+1\right)=0\)
⇔3x+1=0
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
h) \(2x^3-7x^2+7x-2=0\)
\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy S = {2; 1; \(\frac{1}{2}\)}
i) \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)
Vậy S = {1;-2}
Bạn đưa quá nhiều bài 1 lúc nên người ta giải được cũng chẳng ai muốn giải đâu, vì nhìn vào đã thấy ngộp rồi. Kinh nghiệm là muốn được giải quyết nhanh thì chỉ đăng 2-3 bài 1 lúc thôi
Bài 1:
a/ \(11-\left(2x+3\right)=3\left(x-4\right)\)
\(\Leftrightarrow11-2x-3=3x-12\)
\(\Leftrightarrow5x=20\)
\(\Rightarrow x=4\)
b/ \(5\left(2x-3\right)-4\left(5x-7\right)=19-2x\)
\(\Leftrightarrow10x-15-20x+28=19-2x\)
\(\Leftrightarrow8x=-6\)
\(\Rightarrow x=-\frac{3}{4}\)
c/
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)
\(\Leftrightarrow x=3\)
d/
\(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow79x=158\)
\(\Rightarrow x=2\)
e/
\(\frac{2-6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)
\(\Leftrightarrow4\left(2-6x\right)-2\left(2+3x\right)=140-5\left(6x+3\right)\)
\(\Leftrightarrow0=-121\) (vô lý)
Vậy pt vô nghiệm
f/
\(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow6x=-5\)
\(\Rightarrow x=-\frac{5}{6}\)
2.
A = x2 - 4x + 10 = (x2 - 2.x.2 + 22) + 6 = (x - 2)2 + 6 \(\ge\) 6
( do (x - 2)2 \(\ge\) 0)
Vậy: GTNN của A là 6 (tại x = 2)
B = x2 - x + 1 = (x2 - 2.x.\(\frac{1}{2}\) + \(\frac{1}{4}\)) + \(\frac{3}{4}\) = \(\left(x-\frac{1}{2}\right)^2\) + \(\frac{3}{4}\) \(\ge\) \(\frac{3}{4}\)
Vậy: GTNN của B là \(\frac{3}{4}\) (tại x = \(\frac{1}{2}\) )
C = 2x2 - 8x = 2 (x2 - 4x) = 2(x2 - 2.x.2 + 4) - 8 = 2(x - 2)2 - 8 \(\ge\) -8
Vậy : GTNN của C là -8 (tại x = 2)
Bài 1:
a)
\((x-5)(2x-1)-4x(x+2)=-(x-1)^2-2x(x-3)\)
\(\Leftrightarrow (2x^2-11x+5)-(4x^2+8x)=-(x^2-2x+1)-(2x^2-6x)\)
\(\Leftrightarrow -2x^2-19x+5=-3x^2+8x-1\)
\(\Leftrightarrow x^2-27x+6=0\)
\(\Leftrightarrow (x-\frac{27}{2})^2=\frac{705}{4}\Rightarrow \left[\begin{matrix} x-\frac{27}{2}=\frac{\sqrt{705}}{2}\\ x-\frac{27}{2}=\frac{-\sqrt{705}}{2}\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{27+\sqrt{705}}{2}\\ x=\frac{27-\sqrt{705}}{2}\end{matrix}\right.\)
b)
\((4x-1)-(2x+3)^2-12x(x+3)=1\)
\(\Leftrightarrow 4x-1-(4x^2+12x+9)-(12x^2+36x)=1\)
\(\Leftrightarrow -16x^2-44x-11=0\)
\(\Leftrightarrow 16x^2+44x+11=0\)
\(\Leftrightarrow (4x+\frac{11}{2})^2=\frac{77}{4}\)
\(\Rightarrow \left[\begin{matrix} 4x+\frac{11}{2}=\frac{\sqrt{77}}{2}\\ 4x+\frac{11}{2}=\frac{-\sqrt{77}}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{\sqrt{77}-11}{8}\\ x=\frac{-\sqrt{77}-11}{8}\end{matrix}\right.\)
a) Ta có: \(a\left(m-n\right)+m-n\)
\(=a\left(m-n\right)+\left(m-n\right)\)
\(=\left(m-n\right)\left(a+1\right)\)
b) Ta có: \(mx+my+5x+5y\)
\(=m\left(x+y\right)+5\left(x+y\right)\)
\(=\left(x+y\right)\left(m+5\right)\)
c) Ta có: \(ma+mb-a-b\)
\(=m\left(a+b\right)-\left(a+b\right)\)
\(=\left(a+b\right)\left(m-1\right)\)
d) Ta có: \(1-xa-x+a\)
\(=\left(a+1\right)-x\left(a+1\right)\)
\(=\left(a+1\right)\left(1-x\right)\)
e) Ta có: \(\left(a-b\right)^2-\left(b-a\right)\left(a+b\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a-b+a+b\right)\)
\(=2a\left(a-b\right)\)
f) Ta có: \(a\left(a-b\right)\left(a+b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-ab\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-ab-a^2+ab-b^2\right)\)
\(=b^2\cdot\left(a+b\right)\)
g) Ta có: \(3x\left(x+7\right)^2-11x^2\left(x+7\right)+9\left(x+7\right)\)
\(=\left(x+7\right)\left[3x\left(x+7\right)-11x^2+9\right]\)
\(=\left(x+7\right)\left(3x^2+21x-11x^2+9\right)\)
\(=\left(x+7\right)\left(-8x^2+21x+9\right)\)
\(=\left(x+7\right)\left(-8x^2+24x-3x+9\right)\)
\(=\left(x+7\right)\left[-8x\left(x-3\right)-3\left(x-3\right)\right]\)
\(=\left(x+7\right)\left(x-3\right)\left(-8x-3\right)\)
h) Ta có: \(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
i) Ta có: \(2x\left(x-3\right)-3\left(x-3\right)^2\)
\(=\left(x-3\right)\left[2x-3\left(x-3\right)\right]\)
\(=\left(x-3\right)\left(2x-3x+9\right)\)
\(=\left(x-3\right)\left(9-x\right)\)
j) Ta có: \(x\left(x-7\right)+\left(7-x\right)^2\)
\(=x\left(x-7\right)+\left(x-7\right)^2\)
\(=\left(x-7\right)\left(x+x-7\right)\)
\(=\left(x-7\right)\left(2x-7\right)\)
k) Ta có: \(3x\left(x-9\right)^2-\left(9-x\right)^3\)
\(=3x\left(x-9\right)^2+\left(x-9\right)^3\)
\(=\left(x-9\right)^2\cdot\left(3x+x-9\right)\)
\(=\left(x-9\right)^2\cdot\left(4x-9\right)\)
a) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
b) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Leftrightarrow-x-21=0\)
\(\Leftrightarrow x=-21\)
Vậy ...
c) \(5x\left(12x+7\right)-3x\left(2x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-6x^2+15x+100=0\)
\(\Leftrightarrow54x^2+50x+100=0\)
\(\Leftrightarrow54\left(x^2+\frac{25}{27}x+\frac{625}{2916}\right)+\frac{290975}{2916}=0\)
\(\Leftrightarrow54\left(x+\frac{25}{54}\right)^2+\frac{290975}{2916}=0\left(ktm\right)\)
Vậy phương trình vô nghiệm.
d) \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x-5=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
Vậy ...
e) \(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\)
\(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
\(\Leftrightarrow-2x^2=0\)
\(\Leftrightarrow x=0\)
Vậy ...
\(A=x^5-5x^4+5x^3-5x^2+5x-1\)
\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x+3\)
\(=3\)
Ta có :
\(A=x^5-5x^4+5x^3-5x^2+5x-1\)
\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)\(A=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+3\)
\(A=3\)
P/s tham khảo nha
hok tốt
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
a, 5x(x-2) + (2-x)=0
⇔5x(x-2) - (x-2) =0
⇔(x-2)(5x-1)=0
\(\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy....
c, (x3 - x2) - 4x2 + 8x -4 =0
⇔x3 - x2 -4x2 + 8x - 4=0
⇔x2(x-1) - 4x(x-1) +4(x-1) =0
⇔(x-1) (x-2)2=0
⇔\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy...
Phần b cậu có chép sai đề không?
.chỗ đó là giải phương trình hay PTĐTTNT vậy?