2x
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`
1-2x/2x + 2x/2x-1 + 1/2x-4x^2
2x+2x+1+2x+2+...+2x+2020=2x+2024-8
Lời giải:
$2^x+2^{x+1}+2^{x+2}+....+2^{x+2020}=2^{x+2024}-8$
$2^x(1+2+2^2+...+2^{2020})=2^{x+2024}-8$
$2^x(2+2^2+2^3+...+2^{2021})=2^{x+2025}-16$
$\Rightarrow 2^x(2+2^2+2^3+...+2^{2021})- (2^x(1+2+2^2+...+2^{2020}))=2^{x+2025}-16-(2^{x+2024}-8)$
$\Rightarrow 2^x(2^{2021}-1)=2^{x+2025}-2^{x+2024}-8$
$\Rightarrow 2^x(2^{2021}-1)=2^{x+2024}(2-1)-8$
$\Rightarrow 2^{x+2021}-2^x=2^{3+2021}-2^3$
$\Rightarrow x=3$
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
a. ( 5 - 2x ) ( 5x + 2x ) + 2x ( x + 3 ) = 4 - 2x² b. ( 3x - 2 )( -2x) + 5x² = -x( x - 3) c. 7 - ( 3 + 2x ) (2x - 3 ) = ( x + 4 )²
a: Sửa đề: (5-2x)(5+2x)+2x(x+3)=4-2x^2
=>25-4x^2+2x^2+6x=4-2x^2
=>6x+25=4
=>6x=-21
=>x=-7/2
b: (3x-2)(-2x)+5x^2=-x(x-3)
=>-6x^2+4x+5x^2=-x^2+3x
=>4x=3x
=>x=0
c: =>7-(4x^2-9)=x^2+8x+16
=>7-4x^2+9-x^2-8x-16=0
=>-5x^2-8x=0
=>5x^2+8x=0
=>x(5x+8)=0
=>x=0 hoặc x=-8/5
2x+2x+1+2x+2+2x+3+...+2x+2021=22025-8
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
2x+3/2x+1-2x+5/2x+7=1-6x^2+9x-9/(2x+1)(2x+7)
\(\frac{2x+3}{2x+1}-\frac{2x+5}{2x+7}=\frac{1-6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Leftrightarrow\frac{\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x+7\right)}=\frac{1-6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Rightarrow\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x+1\right)=1-6x^2+9x-9\)
\(\Leftrightarrow4x^2+20x+21-4x^2-12x-5=1-6x^2+9x-9\)
\(\Leftrightarrow8x-16=1-6x^2+9x-9\)
\(\Leftrightarrow8x-16-1+6x^2-9x+9=0\)
\(\Leftrightarrow6x^2-x-8=0\)
Tự làm nốt nha
Trl
-Bạn chuyên toán thcs làm đúng r nhé !~
Học tốt
nhé bạn ~
\(\frac{2x+3}{2x+1}-\frac{2x+5}{2x+7}=1-\frac{6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Leftrightarrow\frac{\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x+7\right)}=\frac{\left(2x+7\right)-6x^2-9x+9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Rightarrow4x^2+20x+21-4x^2-12x-5=2x+7-6x^2-9x+9\)
\(\Leftrightarrow8x+16=-6x^2-7x+16\)
\(\Leftrightarrow6x^2+7x+8x=0\)
\(\Leftrightarrow6x^2+15x=0\)
\(\Leftrightarrow x\left(6x+15\right)=0\)
Đến đây tự làm nốt nha
hok tốt
1-2x/2x+2x/2x-1+1/2x-4x^2
1-2x/2x+2x/2x-1+1/2x-4x^2