(4+x).33=35
65-4x-3=2020^0
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a, x=-505
b, x=35/8 hoac -37/8
nhung cau con lai thi tong tu
1. 3x - 36 = 12 . 75 + 25 . 12
3x - 36 = 12 . (75+25)
3x - 36 = 12 . 100
3x - 36 = 1200
3x = 1200 + 36
3x = 1236
=> x = 1236 : 3 = 412
câu 1 thôi nhá bạn
3x - 36 = 12 . 75 + 25 . 12
3x - 36 = 12 . (75 + 25)
3x - 36 = 1200
3x = 1164
x = 388
x : 2 - 12 = 33 . 40 + 33 . 59 + 33
x : 2 - 12 = 33 . 40 + 33 . 59 + 33 . 1
x : 2 - 12 = (40 + 59 + 1)
x : 2 - 12 = 3300
x : 2 = 3288
x = 1644
(x - 4) . (9 - x) = 0
Thỏa mãn điều kiện\(\hept{\begin{cases}x=4\\x=9\end{cases}}\)
(x - 6) . (2020 - x) = 0
Thỏa mãn điều kiện\(\hept{\begin{cases}x=6\\x=2020\end{cases}}\)
x . (6 - x) = 0
Thỏa mãn điều kiện 6 - x = 0
x = 6
(x - 3 - 12) . (20 - x) = 0
Thỏa mãn điều kiện \(x\le20\); 20 - x = 0
x = 20
`2)x^4+2x^3-x^2-2x+1=0`
`<=>x^4+2x^3+x^2-2x^2-2x+1=0`
`<=>(x^2+x)^2-2(x^2+x)+1=0`
`<=>(x^2+x-1)^2=0`
`<=>x^2+x-1=0`
`\Delta=1+4=5`
`=>x_{1,2}=(-1+-sqrt5)/2`
Vậy `S={(-1+sqrt5)/2,(-1+sqrt5)/2`
`3)x^4-4x^3-9x^2+8x+4=0`
`<=>x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0`
`<=>(x-1)(x^3-3x^2-12x-4)=0`
`<=>(x-1)(x^3+2x^2-5x^2-10x-2x-4)=0`
`<=>(x-1)(x+2)(x^2-5x-10)=0`
`+)x=1`
`+)x=-2`
`+)x^2-5x-10=0`
`Delta=25+40=65`
`=>x_{12}=(5+sqrt{65})/2`
Kết quả : Viết lại biểu thức đã cho
=> -7/4x . ( 33/12 + 33/20 + 33/30 + 33/42 ) = 22
-7/4x . 33 . ( 1/12 + 1/20 + 1/30 + 1/42 ) = 22
-231/4x . ( 1/3 . 4 + 1/ 4. 5 + 1/5 . 6 + 1/ 6. 7 ) = 22
-231/4x . ( 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 ) = 22
-231/4x . ( 1/3 - 1/7 ) = 22
-231/4x . 4/21 = 22
-11x = 22
x = 22 : -11
x = -2
Vậy x = -2
4x2 + 2y2 + 2z2 - 4xy + 2yz - 4xz - 6y - 10z + 34 = 0
<=> [ ( 4x2 - 4xy + y2 ) - 4xz + 2yz + z2 ] + ( y2 - 6y + 9 ) + ( z2 - 10z + 25 ) = 0
<=> [ ( 2x - y )2 - 2( 2x - y )z + z2 ] + ( y - 3 )2 + ( z - 5 )2 = 0
<=> ( 2x - y - z )2 + ( y - 3 )2 + ( z - 5 )2 = 0
\(\hept{\begin{cases}\left(2x-y-z\right)^2\\\left(y-3\right)^2\\\left(z-5\right)^2\end{cases}}\ge0\forall x,y,z\Rightarrow\left(2x-y-z\right)+\left(y-3\right)^2+\left(z-5\right)^2\ge0\forall x,y,z\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\)
Thế vào S ta được :
S = ( x - 4 )2020 + ( y - 3 )2020 + ( z - 5 )2020
= ( 4 - 4 )2020 + ( 3 - 3 )2020 + ( 5 - 5 )2020
= 0 + 0 + 0
= 0
Bài 2 :
a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)
b, \(5x\left(x-2020\right)-x+2020=0\)
\(\Leftrightarrow5x\left(x-2020\right)-\left(x-2020\right)=0\Leftrightarrow\left(5x-1\right)\left(x-2020\right)=0\)
\(\Leftrightarrow x=\frac{1}{5};2020\)
c, \(\left(4x+5\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow16x^2+40x+25-\left(4x^2-4x+1\right)=0\)
\(\Leftrightarrow12x^2+44x+24=0\Leftrightarrow4\left(x+3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow x=-3;-\frac{2}{3}\)
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
(4 + x) . 33 = 35
4 + x = 35 : 33
4 + x = 32
4 + x = 9
x = 9 - 4
x = 5
65 - 4x - 3 = 20200
65 - 4x - 3 = 1
65 - 4x = 1 + 4
65 - 4x = 5
4x = 65 - 5
4x = 50
x = 60 : 4
x = 15
(4 + x) . 33 = 35
(4 + x) = 35 : 33
4 + x = 32
4 + x = 9
x = 9 - 4
x = 5
Vậy x = 5