Tìm x (4x+3)^2-(4x=3)^2-5x-2=0 (x+5)^2-16=0 !!!
BT2: Tìm x 2, 3x(x-4)+2x-8=0 3, 4x(x-3)+x^2-9=0 4, x(x-1)-x^2+3x=0 5, x(2x-1)-2x^2+5x=16
2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4
2x^2 – 6x + 1 = 0
4x^2 – 12x + 5 = 0
2x2 + 5x + 3 = 0
x ^2 + x – 2 = 0
x^ 2 – 4x + 3 = 0
2x^2 + 5x – 3 = 0
x^ 2 + 6x – 16 = 0
a,\(2x^2-6x+1=0\)
\(=>x.\left(2x-6\right)=1\)
\(th1:\orbr{\begin{cases}x=1\\2x-6=1\end{cases}=>\orbr{\begin{cases}x=1\\x=\frac{7}{2}\end{cases}}}\)
\(th2:\orbr{\begin{cases}x=-1\\2x-6=-1\end{cases}=>\orbr{\begin{cases}x=-1\\x=\frac{5}{2}\end{cases}}}\)
b,\(4x^2-12x+5=0\)
\(=>x.\left(4x-12\right)=-5\)
\(th1:\orbr{\begin{cases}x=1\\4x-12=-5\end{cases}=>\orbr{\begin{cases}x=1\\x=\frac{7}{4}\end{cases}}}\)
\(th2:\orbr{\begin{cases}x=-1\\4x-12=5\end{cases}=>\orbr{\begin{cases}x=-1\\x=\frac{17}{4}\end{cases}}}\)
\(th3:\orbr{\begin{cases}x=5\\4x-12=-1\end{cases}=>\orbr{\begin{cases}x=5\\x=\frac{11}{4}\end{cases}}}\)
\(th4:\orbr{\begin{cases}x=-5\\4x-12=1\end{cases}=>\orbr{\begin{cases}x=-5\\x=\frac{13}{4}\end{cases}}}\)
\(x^2+6x-16=0\)
Ta có \(\Delta=6^2+4.16=100,\sqrt{\Delta}=10\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-6+10}{2}=2\\x=\frac{-6-10}{2}=-8\end{cases}}\)
\(2x^2+5x-3=0\)
Ta có \(\Delta=5^2+4.2.3=49,\sqrt{\Delta}=7\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-5+7}{4}=\frac{1}{2}\\x=\frac{-5-7}{4}=-3\end{cases}}\)
a) x^2+9x=0
b)x^3+4x^2+4x=0
c)(x-5)^2-16=0
d)3(x+2)-x^2-2x=0
e)(x+2)^3-x^2(x+6)=4
g)x^2+5x+6=0
\(a,x\left(x+9\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Rightarrow x\left(x^2+4x+4\right)=0\\ \Rightarrow x\left(x+2\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(x-5-4\right)\left(x-5+4\right)=0\\ \Rightarrow\left(x-9\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\\ d,\Rightarrow3\left(x+2\right)-x\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ e,\Rightarrow x^3+6x^2+12x+8-x^3-6x^2=4\\ \Rightarrow12x=-4\Rightarrow x=-\dfrac{1}{3}\\ g,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Tìm x :
a) (x+2)^2-(x-2)(x+2)=0
b) (2x+3)^2-(x-1)^2=0
c) x^3-8=(x-2)^2
d) x^3+5x^2-4x-20=0
e) x^3-4x^2+4x=0
f) x^2-25+2(x+5)=0
g) 2(x^2+8x+16)-x^2+4=0
h) x^2(x-2)+7x=14
Mai nộp rồi giúp mình với các bạn
Giai phường trình sau:
a, \(3x^2+2x-1=0\) e, \(4x^2-12x+5=0\) i,\(2x^2+5x-3=0\)
b,\(x^2-5x+6=0\) f, \(2x^2+5x+3=0\) j,\(x^2+6x-16=0\)
c,\(x^2-3x+2=0\) g,\(x^2+x-2=0\)
d,\(2x^2-6x+1=0\) h, \(x^2-4x+3=0\)
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)
(Bài 14; Tìm x biết
1) x ^ 2 - 9 = 0
4) 4x ^ 2 - 4 = 0
7) (3x + I) ^ 2 - 16 = 0
10) (x + 3) ^ 2 - x ^ 2 = 45
2) 25 - x ^ 2 = 0
5) 4x ^ 2 - 36 = 0
8) (2x - 3) ^ 2 - 49 = 0
11) (5x - 4) ^ 2 - 49x ^ 2 = 0
3) - x ^ 2 + 36 = 0
6) 4x ^ 2 - 36 = 0
9) (2x - 5) ^ 2 - x ^ 2 = 0
12) 16 * (x - 1) ^ 2 - 25 = 0
1, \(x^2\) - 9 = 0
(\(x\) - 3)(\(x\) + 3) = 0
\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
vậy \(x\) \(\in\) {-3; 3}
7, (3\(x\) + 1)2 - 16 = 0
(3\(x\) + 1 - 4)(3\(x\) + 1 + 4) = 0
(3\(x\) - 3).(3\(x\) + 5) = 0
\(\left[{}\begin{matrix}3x-3=0\\3x+5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=3\\3x=-5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=\dfrac{-5}{3}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {1; - \(\dfrac{5}{3}\)}
10, (\(x\) + 3)2 - \(x^2\) = 45
[(\(x\) + 3) - \(x\)].[(\(x\) + 3) + \(x\)] = 45
3.(2\(x\) + 3) = 45
2\(x\) + 3 = 15
2\(x\) = 12
\(x\) = 6
Tìm x:
1, x2 - 2x +3=0
2, 5x .(x-1) = x-1
3, 2.(x+5) -x2 -5x=0
4, x+5x2 =0
5, x+1=(x+1)2
6, 5x.(4x-5)=(2x-3).(4x-5)
7, (4x2+2x).(x2 -x)+(4x2 +6).(x-x2)=0
8, (x+12)2 - 9x2=0
1)⇔x2+1x-3x+3=0
⇔x(x+1)-3(x+1)=0
⇔(x+1)(x-3)=0
⇔x+1=0 hoặc x-3=0
⇔x=-1 hoặc x=3
4)⇔x(1+5x)=0
⇔x=0 hoặc 1+5x=0
⇔x=0 hoặc 5x=-1
⇔x=0 hoặc x=-0.2
Tìm x,biết
1) 3x^2 - 4x = 0
2) (x^2 - 5x) + x - 5 = 0
3) x^2 - 5x + 6 = 0
4) 5x(x-3) - x+3 = 0
5) x^2 - 2x + 5 = 0
6) x^2 + x -6 = 0
Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
Tìm x:
a, 3x (4x -3) - 2x (5-6x) = 0
b, 5 (2x-3) + 4x (x-2) + 2x (3-2x) = 0
c, 3x (2-x) + 2x (x-1) = 5x (x+3)
d, 3x (x+1) - 5x (3-x) + 6(x2 + 2x + 3) = 0
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
d) \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
Mà \(14x^2+18>0\)nên pt vô nghiệm
Tìm x
a) 3x(4x - 3) - 2x(5 - 6x) = 0
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
d) 3x (x + 1) - 5x(3 - x) + 6(x^2 + 2x + 3) = 0
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
b) 5(2x-3)+4x(x-2)+2x(3-2x)=0
\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
\(\Leftrightarrow x=\dfrac{15}{8}\)
vậy x=\(\dfrac{15}{8}\)
c)3x(2-x)+2x(x-1)=5x(x+3)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)
\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)