Tìm x biết: {[(9x+3x-128) x 3] x 10} + 1000 =760
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Ta có;
x+2x+3x+4x+...+9x=1000-100
=>x+2x+3x+4x+...+9x=900
=>(1+2+3+4+....+9)x=900
=>[(1+9).9:2]x=900
=>45x=900
x=900:45=20
cho tớ đúng nha
Ta có;
x+2x+3x+4x+...+9x=1000-100
=>x+2x+3x+4x+...+9x=900
=>(1+2+3+4+....+9)x=900
=>45x=900
x=900:45=20
a)
\(512-\left(128-5x\right)=3x+12\\ 512-128+5x=3x+12\\ 384+5x=3x+12\\ 5x-3x=12-384\\ 2x=-372\\ x=\left(-372\right):2\\ x=-186\)
b)
\(\left(2x-1\right)+\left(4x-2\right)+...+\left(400x-200\right)=5+10+...+1000\\ \left(2x+4x+...+400x\right)-\left(1+2+...+200\right)=5+10+...+1000\\ x\left(2+4+...+400\right)=\left(5+10+...+1000\right)+\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=5\cdot\left(1+2+...+200\right)+1\cdot\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=\left(5+1\right)\cdot\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=6\cdot\left(1+2+...+200\right)\\ \Rightarrow2x=6\\ x=6:2\\ x=3\)
c)
\(\left(x+2\right)+\left(4x+4\right)+\left(7x+6\right)+...+\left(25x+18\right)+\left(28x+20\right)=1560\\ \left(x+4x+7x+...+25x+28x\right)+\left(2+4+6+...+18+20\right)=1560\\ x\left(1+4+7+...+25+28\right)+110=1560\\ 145x+110=1560\\ 145x=1560-110\\ 145x=1450\\ x=1450:145\\ x=10\)
d)
\(x+4x+5x+9x+14x+...+97x=500\\ x\left(1+4+5+9+14+...+97\right)=500\)
Dãy số trong ngoặc có quy luật: Số thứ \(n\) bằng số thứ \(n-1\) cộng số thứ \(n-2\)
Suy ra dãy số đó là: \(1+4+5+9+14+23+37+60+97=250\)
Thế vào ta được:
\(250x=500\\ x=500:250\\ x=2\)
e)
\(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\\ 720-41+\left(2x-5\right)=8\cdot5\\ 720-41+2x-5=40\\ \left(720-41-5\right)+2x=40\\ 674+2x=40\\ 2x=40-674\\ 2x=-634\\ x=\left(-634\right):2\\ x=-317\)
f)
\(697:\dfrac{15x+364}{x}=17\\ \dfrac{15x+364}{x}=697:17\\ \dfrac{15x+364}{x}=41\\ \dfrac{15x+364}{x}\cdot x=41x\\ 15x+364=41x\\ 364=41x-15x\\ 364=26x\\ x=364:26\\ x=14\)
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`
\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
a, 5\(x\)(\(x^2\) - 9) = 0
\(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -3; 0; 3}
b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0
3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0
(\(x+3\))( 3 - \(x\)) = 0
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -3; 3}
c, \(x^2\) - 9\(x\) - 10 = 0
\(x^2\) + \(x\) - 10\(x\) - 10 = 0
\(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0
(\(x+1\))(\(x-10\)) = 0
\(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -1; 10}
a) x3 - 19x - 30 = 0
\(\Leftrightarrow\)x3 + 5x2 + 6x - 5x2 - 25x - 30 = 0
\(\Leftrightarrow\)(x - 5)(x2 + 5x + 6) = 0
\(\Leftrightarrow\)(x - 5)(x2 + 2x + 3x + 6) = 0
\(\Leftrightarrow\)(x - 5)(x + 2)(x + 3) = 0
\(\Leftrightarrow\)x - 5 = 0 x = 5
hoặc x + 2 = 0 \(\Leftrightarrow\) x = -2
hoặc x + 3 = 0 x = -3
Vậy x = { -3; -2; 5 }
b) x(x + 4)(x + 6)(x + 10) + 128 = 0
\(\Leftrightarrow\)(x2 + 10x)(x2 + 10x + 24) + 128 = 0
Đặt x2 + 10x = y; ta có
y(y + 24) + 128 = 0
\(\Leftrightarrow\)y2 + 24y + 144 - 16 = 0
\(\Leftrightarrow\)(y + 12)2 - 16 = 0
\(\Leftrightarrow\)(y + 12 - 4)(y + 12 + 4) = 0
\(\Leftrightarrow\)(y + 8)(y + 16) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}y+8=0\\y+16=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y=-8\\y=-16\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2+10x=-8\\x^2+10x=-16\end{cases}}\)
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
a: =>x^2-25-x^2-3x=10
=>-3x=35
=>x=-35/3
b: =>4x^2-9-4(x^2+4x+4)=5
=>4x^2-9-4x^2-16x-16-5=0
=>-16x-30=0
=>x=-15/8
c: =>9x^2+45x-9x^2+4=7
=>45x=3
=>x=1/15
d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8
=>8x=7
=>x=7/8
{[(9x+3x-128) x 3] x 10} + 1000 =760
{[(9x+3x-128) x 3] x 10} =760 - 1000
{[(9x+3x-128) x 3] x 10} =-240
{[(9x+3x-128) x 3] x 10} =-240
{[x(9+3)-128] x 3} =-240 : 10
{[x(9+3)-128] x 3} =-24
{[12x-128] x 3} =-24
[12x-128] =-24 : 3
[12x-128] =-8
12x =-8+128
12x =120
x =120:12
x =10
{[(9x+3x-128) x 3] x 10} + 1000 =760
[(x(9+3)-128)x3]x10=760-1000
[(x.12-128)x3]x10=-240
(x.12-128)x3=-240:10
(x.12-128)x3=-24
(x.12-128)=-24:3
x.12-128=-8
x.12=-8+128
x.12=120
x=120:12
x=10
vậy x=10