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a) ( 6x + 1 )2 + ( 6x - 1 )2 - 2( 1 + 6x )( 6x - 1 )
= ( 6x + 1 )2 - 2( 1 + 6x )( 6x - 1 ) + ( 6x - 1 )2
= ( 6x + 1 - 6x + 1 )2 = 22 = 4
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
2.$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
$\textbf{5)}$
$(4x-5)(x+2)-(x+5)(x-3)-3x^2-x$
$=(4x^2+3x-10)-(x^2+2x-15)-3x^2-x$
$=4x^2+3x-10-x^2-2x+15-3x^2-x$
$=5.$
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+12x+1+36x^2-12x+1-72x^2+2\)
\(=4\)
c) \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
$(6x+1)^2-(6x-1)-2(1+6x)(6x-1)$
$=36x^2+12x+1-6x+1-2(6x+1)(6x-1)$
$=36x^2+6x+2-2(36x^2-1)$
$=36x^2+6x+2-72x^2+2$
$=-36x^2+6x+4$
b)$x(2x^2-3)-x^2(5x+1)+x^2$
$=2x^3-3x-5x^3-x^2+x^2$
$=2x^3-5x^3-3x$
$=-3x^3-3x$
$=-3x(x^2+1)$
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$
d)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)
\(=6x^2+12x+x+2-6x^2+10x\)
\(=23x+2\)
a) (6x + 1)(x + 2) - 2x(3x - 5)
= 6x2 + 12x + x + 2 - 6x2 + 10x
= (6x2 - 6x2) + (12x + x + 10x) + 2
= 23x + 2
b) (2x - 1)2 - (2x - 3)(2x + 3)
= 4x2 - 4x + 1 - 4x2 + 9
= (4x2 - 4x2) - 4x + (1 + 9)
= -4x + 10
c) (2x - 3)3 - (3x + 1)(5 - 4x) - 16x2
= 8x3 - 36x2 + 54x - 15x + 12x2 - 5 + 4x - 16x2
= 8x3 - (36x2 - 12x2 + 16x2) + (54x - 15x + 4x) - 5
= 8x3 - 40x2 + 43x - 5
d) (3x + 2) - (x - 5) - x(3x - 13)
= 3x + 2 - x + 5 - 3x2 + 13x
= (3x - x + 13x) + (2 + 5) - 3x2
= 15x + 7 - 3x2
$\textbf{1)}$
$(4x-1)^2-2(4x-1)(3x-7)+(7-3x)^2$
$=(4x-1)^2-2(4x-1)(3x-7)+(3x-7)^2$
$=\left[(4x-1)-(3x-7)\right]^2$
$=(x+6)^2.$
Tại $x=44$: $(44+6)^2=50^2=2500.$
$\textbf{2)}$
$(2x-5)^2-2(2x-5)(3x-4)+(4-3x)^2$
$=(2x-5)^2-2(2x-5)(3x-4)+(3x-4)^2$
$=\left[(2x-5)-(3x-4)\right]^2$
$=(x+1)^2.$
Tại $x=24$: $(24+1)^2=625.$
\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
\(=4\)
\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=\left(2x^3-5x^3\right)+\left(x^2-x^2\right)-3x\)
\(=-3x^3-3x\)
\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)-\left(6x+5x\right)+24\)
\(=-11x+24\)
1: \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
2: \(=8x^2+12x-10x-15-4\left(2x^2-x+4x-2\right)+10x+7\)
\(=8x^2+12x-8-8x^2+4x-16x+8\)
\(=0\)
3: \(=\left(6x+1-6x+1\right)^2=4\)
5: \(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)
\(=3x^3+6x-3x^3+3x=9x\)
$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=24-11x$
2.$(4x-5)(2x+3)-4(x+2)(2x-1)+10x+7$
$=8x^2+12x-10x-15-4(2x^2+3x-2)+10x+7$
$=8x^2+2x-15-8x^2-12x+8+10x+7$
$=(8x^2-8x^2)+(2x-12x+10x)+(-15+8+7)$
$=0$
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
4.$(x^2-2x+2)(x^2-2)(x^2+2x+2)(x^2+2)$
$=[(x^2+2)-2x](x^2+2)+2x(x^2+2)$
$=(x^2+2)^2-(2x)^2$
$=(x^4+4x^2+4-4x^2)(x^4-4)$
$=(x^4+4)(x^4-4)$
$=x^8-16$
$(x+1)^3+(x-1)^3+x^3-3x(x+1)(x-1)$
$=(x^3+3x^2+3x+1)+(x^3-3x^2+3x-1)+x^3-3x(x^2-1)$
$=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x^3+3x$
$=(x^3+x^3+x^3-3x^3)+(3x^2-3x^2)+(3x+3x+3x)+(1-1)$
$=9x$
6.$3(2^2+1)(2^4+1)(2^8+1)\cdots(2^{64}+1)+1$
Ta có:
$(a-1)(a+1)=a^2-1$
Nên:
$(2^2-1)(2^2+1)=2^4-1$
$(2^4-1)(2^4+1)=2^8-1$
$(2^8-1)(2^8+1)=2^{16}-1$
$\ldots$
$(2^{32}-1)(2^{32}+1)=2^{64}-1$
Do đó:
$(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)=2^{64}-1$
Vì $2^2-1=3$, nên:
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)=2^{64}-1$
Suy ra:
$3(2^2+1)(2^4+1)(2^8+1)\cdots(2^{64}+1)+1$
$=2^{64}$