PT ⇔ \(4x^2-24x+36=x^2-6x+9\)
⇔ \(3x^2-18x+27=0\)
⇔ \(\left(3x-9\right)\left(x-3\right)=0\)⇔ \(x=3\)
PT ⇔ \(4x^2-24x+36=x^2-6x+9\)
⇔ \(3x^2-18x+27=0\)
⇔ \(\left(3x-9\right)\left(x-3\right)=0\)⇔ \(x=3\)
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, \(\left(x+3\right)^4-\left(x-3\right)^4-24x^2\left(x-1\right)=108\)
b, \(\left(x+2\right)^5-\left(x-2\right)^5=108\)
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)\(12x-x^2-36=0\)
Tìm x, biết:
a) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)
b) \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
Rút gọn
\(\frac{\left(4x-3\right)^2-9x^2}{-7x^2+24x-9}\)
Cho biểu thức
A= \(\left[\frac{3}{2}\left(x^4-\frac{x^4+1}{x^2+1}\right).\frac{x^3-x\left(4x-1\right)-4}{x^7+6x^6-x-6}\right]:\frac{x^2+29x+78}{3x^2+12x-36}\)
a) Rút gọn A
b) Tìm x nguyên để A có giá trị nguyên
1) Cho P = \(\left(\dfrac{4x-x^3}{1-4x^2}-x\right):\left(\dfrac{4x^2-x^4}{1-x^2}+1\right)\)
a) rút gọn b) tìm x để P > 0
2) Cho Q = \(\left(\dfrac{x}{x^2-3x+9}-\dfrac{11}{x^3+27}+\dfrac{1}{x+3}\right):\dfrac{x^2-1}{x+3}\)
a) rút gọn b) tìm GTLN
3) Cho A = \(\dfrac{1}{\left(x-y\right)^3}\left(\dfrac{1}{x^3}-\dfrac{1}{y^3}\right)+\dfrac{3}{\left(x-y\right)^4}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{6}{\left(x-y\right)^5}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
chứng minh A là lập phương một số hữu tỉ
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}
Bài 1 : dùng hẳng đẳng thức để khai triển và thu gọn
a) \(\left(2x^2+\frac{1}{3}\right)^3\)
b) \(\left(2x^2y-3xy\right)^3\)
c) \(\left(-3xy^4+\frac{1}{2}x^2y^2\right)^3\)
d) \(\left(-\frac{1}{3}ab^2-2a^3b\right)^3\)
e) \(\left(x+1\right)^3-\left(x-1\right)^3-6.\left(x-1\right).\left(x+1\right)\)
f) \(x.\left(x-1\right).\left(x+1\right)-\left(x+1\right).\left(x^2-x+1\right)\)
g) \(\left(x-1\right)^3-\left(x+2\right).\left(x^2-2x+4\right)+3.\left(x-4\right).\left(x+4\right)\)
h) \(3x^2.\left(x+1\right).\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right).\left(x^4+x^2+1\right)\)
k) \(\left(x^4-3x^2+9\right).\left(x^2+3\right)+\left(3-x^2\right)^3-9x^2.\left(x^2-3\right)\)
l) \(\left(4x+6y\right).\left(4x^2-6xy+9y^2\right)-54y^3\)