\(\sqrt{1+2020x}+\sqrt{1-2020x}=\left(1+2021x\right)\sqrt{1-2021x}+\left(1-2021x\right)\sqrt{1+2021x}\)
Những câu hỏi liên quan
Rút gọn:
a) A=(5-2x)2-4x(x-5)
b) B= (4-3x)(4+3x)+(3x+1)2
c) C= (x+1)3-x(x2+3x+3)
d) D=(2021x-2020)2-2(2021x-2020)(2020x-2021)+(2020x-2021)
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
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Tìm x biết:2021-x+2021x(1-2020x)=0
2021 - x + 2021(x - 2020x) = 0
<=> 2021 - x + 2021 - 4082420 = 0
<=> -x - 4082420 = 0
<=> x = -4082420
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Giải phương trình:
\(\dfrac{1}{2018x+1}-\dfrac{1}{2019x+2}=\dfrac{1}{2020x+4}-\dfrac{1}{2021x+5}\)
ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2018};-\dfrac{2}{2019};-\dfrac{1}{505};\dfrac{-5}{2021}\right\}\)
Ta có: \(\dfrac{1}{2018x+1}-\dfrac{1}{2019x+2}=\dfrac{1}{2020x+4}-\dfrac{1}{2021x+5}\)
\(\Leftrightarrow\dfrac{2019x+2-2018x-1}{\left(2018x+1\right)\left(2019x+2\right)}=\dfrac{2021x+5-2020x-4}{\left(2020x+4\right)\left(2021x+5\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(2018x+1\right)\left(2019x+2\right)}=\dfrac{x+1}{\left(2020x+4\right)\left(2021x+5\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(2018x+1\right)\left(2019x+2\right)}-\dfrac{x+1}{\left(2020x+4\right)\left(2021x+5\right)}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{\left(2018x+1\right)\left(2019x+2\right)}-\dfrac{1}{\left(2020x+4\right)\left(2021x+5\right)}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\dfrac{1}{\left(2018x+1\right)\left(2019x+2\right)}=\dfrac{1}{\left(2020x+4\right)\left(2021x+5\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(2018x+1\right)\left(2019x+2\right)=\left(2020x+4\right)\left(2021x+5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\4074342x^2+6055x+2=4082420x^2+18184x+20\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\-8078x^2-12129x-18=0\end{matrix}\right.\)
Ta có: \(-8078x^2-12129x-18=0\)(2)
\(\Delta=\left(-12129\right)^2-4\cdot\left(-8078\right)\cdot\left(-18\right)=146531025\)
Vì \(\Delta>0\) nên phương trình (2) có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{12129-12105}{2\cdot\left(-8078\right)}=\dfrac{-6}{4039}\left(nhận\right)\\x_2=\dfrac{12129+12105}{2\cdot\left(-8078\right)}=-\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{-6}{4039};\dfrac{-3}{2}\right\}\)
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Cho x= 1+\(\sqrt{2021}\). Tính giá trị biểu thức
Q= \(x^5-2x^4-2021x^3+3x^2+2018x-2021\)
Cho f(x)=x^2014-2021x^2013+2021x^2012-...+2021x^2-2021x-1
Tính f(2020)
Giúp nhanh nha mình ticks cho
Xem chi tiết
Ta có: a = 2020 => 2021 = x + 1
f(2020) = x2014 - (x + 1) . x2013 + (x + 1) . x2012 - ... + (x + 1) . x2 - (x + 1) . x - 1
= x2014 - x2014 + x2013 + x2013 + x2012 - ... + x3 + x2 - x2 + x - 1
= x - 1 = 2020 - 1 = 2019
Vậy f(2020) = 2019
Các bạn giúp mình với ạ!
Cho x = 2020, tính giá trị của biểu thức: x^2020 - 2021x^2019 + 2021x^2018 - 2021x^2017 + ... + 2021x^2 - 2021x +1
x = 2020 => 2021 = x + 1
x2020 - 2021x2019 + 2021x2018 - 2021x2017 + ... + 2021x2 - 2021x + 1
= x2020 - ( x + 1 )x2019 + ( x + 1 )x2018 - ( x + 1 )x2017 + ... + ( x + 1 )x2 - ( x + 1 )x + 1
= x2020 - x2020 - x2019 + x2019 + x2018 - x2018 - x2017 + ... + x3 + x2 - x2 - x + 1
= -x + 1 = -2020 + 1 = -2019
Vậy giá trị của biểu thức = -2019
Phân tích đa thức thành nhân tử
x^4+2021x^2-2020x+2021
x4 + 2021x2 - 2020x + 2021
= (x4 + x) + 2021(x2 - x + 1)
= x(x3 + 1) + 2021(x2 - x + 1)
= x(x + 1)(x2 - x + 1) + 2021(x2 - x + 1)
= (x2 + x + 2021)(x2 - x + 1)
giải phương trình:\(\left(1+\sqrt{x^2+2020x}+2019\right)\left(\sqrt{x+2019}-\sqrt{x+1}\right)=2018\)
giải phương trình sau:\(\left(1+\sqrt{x^2+2020x}+2019\right)\left(\sqrt{x+2019}-\sqrt{x+1}\right)=2018\)
