|2x-1|<2x-3
Giải bất pt chứa dấu gttđ
Lớp 10
1) Đa thức \(4x^2+1\) được phân tích thành nhân tử là:
A)\(\left(2x^2-2x-1\right)\left(2x^2+2x-1\right)\)
B)\(\left(2x^2+2x+1\right)\left(2x^2+2x-1\right)\)
C)\(\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
D)\(\left(2x^2+2x+1\right)\left(2x^2-2x-1\right)\)
2) Đa thức \(4x^4+y^4\) được phân tích thành nhân tử là:
A)\(\left(2x^2+2xy+y^2\right)\left(2x^2+2xy-y^2\right)\)
B)\(\left(2x^2+2xy-y^2\right)\left(2x^2-2xy+y^2\right)\)
C)\(\left(2x^2+2xy+y^2\right)\left(2x^2-2xy+y^2\right)\)
D) Một kết quả khác
`1)4x^2+1=4x^2+4x+1-4x=(2x+1)^2-4x=(2x-2\sqrt{x}+1)(2x+2\sqrt{x}+1)` (với `x >= 0`)
`->` Ko có đ/á
(Câu này mình nghĩ là `4x^4+1` chứ nhỉ?)
`2)4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2`
`=(2x^2+y^2)-(2xy)^2`
`=(2x^2-2xy+y^2)(2x^2+2xy+y^2)`
`->bb C`
cho M= (√x+1√2x+1+√2x+√x√2x−1−1)÷(1+√x√2x+1−√2x+√x√2x−1)
Bạn ghi lại đề đi bạn, khó nhìn quá
căn 2 ( 1-2x) +căn 2 (1+2x) =căn 2 (1-2x/1+2x)+căn 2 (1+2x/1-2x)
a, 2x( 2x-1) -(2x-1)
b, 2x( 4x + 2x + 1) - ( 4x + 2x +1)
a)2x( 2x-1) -(2x-1)
=(2x-1)(2x-1)
=(2x-1)2
b)2x( 4x + 2x + 1) - ( 4x + 2x +1)
=(2x-1)(4x+2x+1)
=(2x-1)(6x+1)
a) \(2x\left(2x-1\right)-\left(2x-1\right)=\left(2x-1\right)\left(2x-1\right)\)
b) \(2x\left(4x+2x+4\right)-\left(4x+2x+4\right)=\left(2x-1\right)\left(4x+2x+4\right)\)
Mik làm cho vui thôi chứ chẳng ai T mik đâu
P=\(\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}-\sqrt{x}}{\sqrt{2x}-1}\right)\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}+\frac{\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{MTC}-\frac{2x-1}{MTC}\)
=\(\frac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{MTC}\)
=\(\frac{2x\sqrt{2}+2\sqrt{2x}}{MTC}\)
1-2x/2x + 2x/2x-1 + 1/2x-4x^2
Tìm x :\(\sqrt{1-2x}+\sqrt{1+2x}=\sqrt{\frac{1-2x}{1+2x}}+\sqrt{\frac{1+2x}{1-2x}}\)
(\(\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}\)-\(\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\)-1):(1+\(\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}\)_\(\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\))
Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
1-2x/2x+2x/2x-1+1/2x-4x^2