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LV
18 tháng 1 2022
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
PN
1 tháng 8 2020
\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)
\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)
\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)
\(=-165+22-108=22-273=-251\)
\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)
\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)
\(=3-7.0+5.\left(-3\right)=3-15=-12\)
\(H=\left(-3x+5\right)\left(x-6\right)-\left(x-1\right)\left(x^2-2x+3\right)+\left(x+2\right)\left(x^2-3\right)\)
\(=\left(3+5\right)\left(-1-6\right)-\left(-1-1\right)\left(1+2+3\right)+\left(-1+2\right)\left(1-3\right)\)
\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)
PN
1 tháng 8 2020
\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)
\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)
\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)
\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)
\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)
\(M=-7x\left(x-5\right)-\left(x-1\right)\left(x^2-x-2\right)+x^2\left(x-3\right)-5x\left(x-8\right)\)
\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)
\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)
\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)
PN
17 tháng 8 2020
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
TT
6 tháng 4 2021
\(a. 2x(3x^2-5x+3) = 6x^3-10x^2+6x \)
\(b. -2x(x^2+5x-3) = -2x^3-10x^2+6x\)
c. \(-\dfrac{1}{2}x^2\left(2x^3-4x+3\right)
=-x^5+2x^3-\dfrac{3}{2}x^2\)
\(d.\left(2x-1\right)\left(x^2+5-4\right)=\left(2x-1\right)\left(x^2+1\right)=2x^3+2x-x^2-1\)
e. \(-\left(5x-4\right)\left(2x+3\right)=10x^2+15x-8x-12=-10x^2+7x-12\)
f.\(\left(2x-y\right)\left(4x^2-2xy+y^2\right)=\left(2x-y\right)\left(2x-y\right)^2=\left(2x-y\right)^3\)
g.\(\left(3x-4\right)\left(x+4\right)+\left(5-x\right)\left(2x^2+3x-1\right)=3x^2+12x-4x-16+10x^2+15x-5-2x^3-3x^2+x=-2x^3+10x^2+24x-21\)
e. \(7x\left(x-4\right)-\left(7x+3\right)\left(2x^2-x+4\right)=7x^2-28x-14x^3+7x^2-28x-6x^2+3x+-12=-14x^3+8x^2-53x-12\)
1 tháng 2 2017
a) x3-x2-21x+45=0
<=> x3+5x2-6x2-30x+9x+45=0
<=> (x+5)(x2-6x+9)=0
<=> (x+5)(x2-3x-3x+9)=0
<=> (x+5)(x-3)2=0
Vậy S={-5;3}
b) X3+3X2+4X+2=0
<=> X3+X2+2X2+2X+2X+2=0
<=> (X+1)(X2+2X+2)=0
VÌ X2+2X+2 >=0
NÊN S={-1}
C) X4+7X-8=0
<=> X4-X3+X3-X2+X2-X+8X-8=0
<=> (X-1)(X3+X2+X+8)=0
VÌ X3+X2+X+8>=0
NÊN S={1}
D) 6X4-X3-7X2+X+1=0
<=> 6X4-6X3+5X3-5X2-2X2+2X-X+1=0
<=> (X-1)(6X3+5X2-2X-1)=0
<=> (X-1)(6X3-3X2+8X2-4X+2X-1)=0
<=> (X-1)(2X-1)(3X2_4X+1)=0
<=> (X-1)(2X-1)(3X2-3x-x+1)=0
<=> (X-1)2(2X-1)(3x-1)=0
vậy S={1/3;1/2;1}
28 tháng 6 2021
a) (7x - 8)(7x + 8) - 10(2x + 3)2 + 5x(3x - 2)2 - 4x(x - 5)2
= 49x2 - 64 - 10(4x2 + 12x + 9) + 5x(9x2 - 12x + 4) - 4x(x2 - 10x + 25)
= 49x2 - 64 - 40x2 - 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x - 100x
= 41x3 - 51x2 - 160x - 154
b) (x2 - 3)(x2 + 3) - 5x2(x + 1)2 - (x2 - 3x)(x2 - 2x) + 4x(x + 2)2
= x4 - 9 - 5x2(x2 + 2x + 1) - x4 + 5x3 - 6x2 + 4x(x2 + 4x + 4)
= 5x3 - 6x2 - 5x4 - 10x3 - 5x2 + 4x3 + 16x2 + 16x - 9
= -5x4 - x3 + 5x2 + 16x - 9
QA
28 tháng 6 2021
Trả lời:
a , ( 7x - 8 ) ( 7x + 8 ) - 10 ( 2x + 3 )2 + 5x ( 3x - 2 )2 - 4x ( x - 5 )2
= 49x2 - 64 - 10 ( 4x2 + 12x + 9 ) + 5x ( 9x2 - 12x + 4 ) - 4x ( x2 - 10x + 25 )
= 49x2 - 64 - 40x2 + 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x2 - 100x
= 41x3 - 11x2 + 40x - 154
b , ( x2 - 3 ) ( x2 + 3 ) - 5x2 ( x + 1 )2 - ( x2 - 3x ) ( x2 - 2x ) + 4x ( x + 2 )2
= x4 - 9 - 5x2 ( x2 + 2x + 1 ) - ( x4 - 2x3 - 3x3 + 6x2 ) + 4x ( x2 + 4x + 4 )
= x4 - 9 - 5x4 - 10x3 - 5x2 - x4 + 2x3 + 3x3 - 6x2 + 4x3 + 16x2 + 16x
= - 5x4 - x3 + 5x2 + 16x - 9