TH1: `5x>=0 <=>x>=0`
`5x-3x=2`
`x=1` (TM)
TH1: `x<0`
`-5x-3x=2`
`x=-1/4` (TM)
Vậy `x=1; x=-1/4`.
TH1: `5x>=0 <=>x>=0`
`5x-3x=2`
`x=1` (TM)
TH1: `x<0`
`-5x-3x=2`
`x=-1/4` (TM)
Vậy `x=1; x=-1/4`.
Thực hiện phép chia
\(\left(5x^6-3x^3+x^2\right):\left(3x^2\right)^2\)
\(a,\left|3x-1\right|=\left|5x-3x\right|\\ b,\left|2x-1\right|+x=2\)
a,\(\left|x+2\right|=\left|-3x+1\right|\)
b,\(\left|x-1,5\right|-\left|3x+2\right|=0\)
c,\(\left|-2+5x\right|=\left|-x^2-2\right|\)
Tìm x biết
1) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
2)\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x+1\right)-33\)
3)\(6x\left(3x+5\right)-2x\left(9x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)-17x^2=0\)
4)\(\left(x-1\right)\left(x+2\right)-\left(x-3\right)+5x-7=0\)
Giúp mình nha. Camon nhiều
\(5x-9=5+3x;2^3+0,5x=1,5;\left(5x+1\right)^2=\dfrac{36}{49};\left(\dfrac{-3}{81}\right)^x=-27;2^{x-1}=16\)
Tìm x biết:
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x-1\)-1
\(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
\(\left|x+1\right|-\left|3x+2\right|=x+2\)
Tìm x biết:
a) \(\left|5x-4\right|=\left|x+2\right|\)
b) \(\left|2+3x\right|=\left|4x-3\right|\)
c) \(\left|2x-3\right|-\left|3x+2\right|=0\)
Tìm x thỏa:
a. \(\left(5x+1\right)^2\)\(-\left(5x+3\right)\)\(\left(5x-3\right)\)= 30
b. \(\left(x+3\right)\)\(\left(x^2-3x+9\right)\)\(-x\left(x-2\right)\left(x+2\right)\)= 15
Ai đúng mik tk !!!
Bài `1`: Rút gọn các biểu thức sau:
\(a)4x^2\left(5x^2+3\right)-6x\left(3x^3-2x+1\right)-5x^3\left(2x-1\right)\)
\(b)\dfrac{3}{2}x\left(x^2-\dfrac{2}{3}x+2\right)-\dfrac{5}{3}x^2\left(x+\dfrac{6}{5}\right)\)
Bài `2`: Thực hiện các phép nhân sau:
\(a)\left(x^2-x\right)\cdot\left(2x^2-x-10\right)\)
\(b)\left(0,2x^2-3x\right)\cdot5\left(x^2-7x+3\right)\)
\(c)6x^2\cdot\left(2x^3-3x^2+5x-4\right)\)
\(d)\left(-1,2x^2\right)\cdot\left(2,5x^4-2x^3+x^2-1,5\right)\)