Có làm theo hàng đẳng thức k bạn?
Có làm theo hàng đẳng thức k bạn?
Bài 3: Tìm x, biết:
a, \(x^3-16x=0\)
b, \(x^4-2x^3+10x^2-20x=0\)
c, \(\left(2x-3\right)^2=\left(x+5\right)^2\)
d, \(x^2\left(x-1\right)-4x^2+8x-4=0\)
tìm x biết
a) \(x^2-2x-3=0\)
b) \(2x^2+5x-3=0\)
c) \(x^4-27x=0\)
d) \(2x\left(x-3\right)-\left(3-x\right)=0\)
e)\(x^2-9=2\left(x+3\right)^2\)
f)\(4x^2-4x+1=\left(5-x\right)^2\)
g) \(4x^2-8x+4=2\left(1-x\right)\left(1+x\right)\)
Bài 3 : Giải các phương trình sau bằng cách đưa về dạng ax+b=0 :
a) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
b) \(4\left(0,5-1,5x\right)=-\frac{5x-6}{3}\)
c) \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
d) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
e) \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right)-\frac{1}{3}\left(x+2\right)\)
tìm x biết
a)5(x+3)-2x(x+3)=0
b)6x\(\left(x^2-2\right)-\left(2-x^2\right)=0\)
c)\(\left(x+1\right)^2-\left(x+1\right)\left(x-2\right)=0\)
d)\(4x\left(x-2017\right)-x+2017=0\)
e)\(\left(x+4\right)^2-16=0\)
f)\(12x-x^2-36=0\)
Giải các phương trình sau:
a) \(x^2+\dfrac{2x}{x-1}=8\)
b) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
c) \(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)
d) \(\left(x^2+6x+8\right)\left(x^2+8x+15\right)=24\)
e) \(\left(x^2+x-2\right)\left(x^2+9x+18\right)=28\)
f) \(3\left(-x^2+2x+3\right)^4-26x^2\left(-x^2+2x+3\right)^2-9x^4=0\)
g) \(x^4+6x^3+11x^2+6x+1=0\)
h) \(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=0\)
i) \(\left(x+2\right)^4+\left(x+8\right)^4=272\)
Giải các phương trình sau
e) \(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3-\frac{1}{3}\left(x+2\right)\)
f)(4-3x)(10x-5)=0
g) (x-3)(2x-1)=(2x-1)(2x+3)
h) 9 - x^2 = 0
Tìm x, biết:
a) \(\left(2x-1\right)^2-4x^2+1=0\)
b) \(6x^3-24x=0\)
c) \(2x\left(x-3\right)-4x+12=0\)
d) \(x^3-5x^2+x-5=0\)
Tìm x
a) x^3 - 16x = 0
b) x^4 - 2x^3 + 10x^2 - 20x = 0
c) (2x - 3 )^2 = (x+5)^2
d) x^2(x-1) - 4x^2 + 8x - 4 = 0
e) x^2 + 4x + 3 = 0
f) x^3 - x^2 = 4x^2 - 8x + 4
g) 2(x+3) - x^2 - 3x = 0
bài 1: khoanh tròn vào chỗ sai trong các bài giải sau và sửa lại cho đúng
a) \(\left(2x+5\right)\left(5-2x\right)=2x^2-5^2\)
b) \(A=\left(x-5\right)^2+\left(2x+1\right)^2-2\left(2x^2+8.5\right)\)
\(A=\left(x^2-10x+25\right)+\left(2x^2+4x+1\right)-4x-17\)
\(A=x^2-6x+9\)
c) \(4x^2=36x-81\)
\(\Leftrightarrow4x^2-36=-81\)
\(\Leftrightarrow4x^2-36+81=0\)
\(\Leftrightarrow\left(2x-9\right)^2=0\)
\(\Leftrightarrow2x-9=0\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{9}{2}\)
vậy S={4,5}
d)\(\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
\(\Leftrightarrow x^2-5-x-3\)
\(\Leftrightarrow x^2-5-x+3=0\)
\(\Leftrightarrow x^2-2-x=0\)
\(\Leftrightarrow x^2-2x+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\) x=0 hoặc x=2
vậy S={0;2}