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Bài 1:
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-2\left(36x^2-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-72x^2+2\)
\(=4\)
b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
c) \(x\left(2x^3-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^4-3x-5x^3-x^2+x^2\)
\(=2x^4-5x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
d, (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = t ta được:
t2 + 3xt + 2x2 = 0
\(\Leftrightarrow\) t2 + xt + 2xt + 2x2 = 0
\(\Leftrightarrow\) t(t + x) + 2x(t + x) = 0
\(\Leftrightarrow\) (t + x)(t + 2x) = 0
Thay t = x2 + 4x + 8 ta được:
(x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
\(\Leftrightarrow\) (x2 + 5x + 8)[x(x + 4) + 2(x + 4)] = 0
\(\Leftrightarrow\) (x2 + 5x + \(\frac{25}{4}\) + \(\frac{7}{4}\))(x + 4)(x + 2) = 0
\(\Leftrightarrow\) [(x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\)](x + 4)(x + 2) = 0
Vì (x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy S = {-4; -2}
Mình giúp bn phần khó thôi!
Chúc bn học tốt!!
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$b) $ \(\left(-x\right)^7:\left(-x\right)^5\)
\(=\left(-x\right)^{7-5}\)
\(=\left(-x\right)^2\)
\(=x^2\)
$c)$ \(5x^2y^4:10x^2y\)
\(=\dfrac{1}{2}y^3\)
$e)$ \(\left(-xy\right)^{14}:\left(-xy\right)^7\)
\(=\left(-xy\right)^{14-7}\)
\(=\left(-xy\right)^7\)
Các câu còn lại tương tự nha bạn!
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
cảm ơn bạn đã nhắc
cmm
$x^2(x-2x^3)$
$=x^3-2x^5$
b)$(x^2+1)(5-x)$
$=5x^2-x^3+5-x$
$=-x^3+5x^2-x+5$
c)$(x-2)(x^2+3x-4)$
$=x^3+3x^2-4x-2x^2-6x+8$
$=x^3+x^2-10x+8$
d)$(x-2)(x-x^2+4)$
$=x^2-x^3+4x-2x+2x^2-8$
$=-x^3+3x^2+2x-8$
e)$(x^2-1)(x^2+2x)$
$=x^4+2x^3-x^2-2x$
f)$(2x-1)(3x+2)(3-x)$
$(3x+2)(3-x)=-3x^2+7x+6$
$=(2x-1)(-3x^2+7x+6)$
$=-6x^3+17x^2+5x-6$
$(x-2y)^2$
$=x^2-4xy+4y^2$
b)$(2x^2+3)^3$
$=(2x^2)^3+3(2x^2)^2\cdot3+3(2x^2)\cdot3^2+3^3$
$=8x^6+36x^4+54x^2+27$
c)$(x-2)(x^2+2x+4)$
$=x^3-8$
d)$(2x-1)^3$
$=(2x)^3-3(2x)^2+3(2x)-1$
$=8x^3-12x^2+6x-1$
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=(6x+1-(6x-1))^2$
$=2^2$
$=4$
b)$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
Dùng:
$(a-1)(a+1)=a^2-1$
Ta có:
$(2^2-1)(2^2+1)=2^4-1$
$(2^4-1)(2^4+1)=2^8-1$
$(2^8-1)(2^8+1)=2^{16}-1$
$(2^{16}-1)(2^{16}+1)=2^{32}-1$
Do đó:
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3(2^{32}-1)$
$=3(4294967296-1)$
$=12884901885$
c)$x(2x^2-3)-x^2(5x+1)+x^2$
$=2x^3-3x-5x^3-x^2+x^2$
$=-3x^3-3x$
$=-3x(x^2+1)$
d)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=24-11x$
$101^2$
$=(100+1)^2$
$=10000+200+1$
$=10201$
b)$97\cdot103$
$=(100-3)(100+3)$
$=100^2-3^2$
$=10000-9$
$=9991$
c)$77^2+23^2+77\cdot46$
$=77^2+23^2+2\cdot77\cdot23$
$=(77+23)^2$
$=100^2$
$=10000$
d)$105^2-5^2$
$=(105-5)(105+5)$
$=100\cdot110$
$=11000$
e)$A=(x-y)(x^2+xy+y^2)+2y^3$
Ta có:
$(x-y)(x^2+xy+y^2)=x^3-y^3$
$A=x^3-y^3+2y^3$
$=x^3+y^3$
Tại $x=\dfrac23,\ y=\dfrac13$:
$A=\left(\dfrac23\right)^3+\left(\dfrac13\right)^3$
$=\dfrac{8}{27}+\dfrac{1}{27}$
$=\dfrac{9}{27}$
$=\dfrac13$