Tính giá trị biểu thức \(P=\frac{\left(2016^2\cdot2026+31\cdot2017-1\right)\left(2016\cdot2021+4\right)}{2017\cdot2018\cdot2019\cdot2020\cdot2021}\)
tính :P=\(\dfrac{\left(2016^2\cdot2026+31\cdot2017-1\right)\left(2016\cdot2021+4\right)}{2017\cdot2018\cdot2019\cdot2020\cdot2021}\)
Đặt \(2016=a\) biểu thức trên trở thành:
\(P=\dfrac{\left(a^2\left(a+10\right)+31\left(a+1\right)-1\right)\left(a\left(a+5\right)+4\right)}{\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)}=\dfrac{A}{B}\)
Với \(B=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)\)
Ta có: \(a^2\left(a+10\right)+31\left(a+1\right)-1=a^3+10a^2+31a+30\)
\(=a^3+5a^2+6a+5a^2+25a+30=a\left(a^2+5a+6\right)+5\left(a^2+5a+6\right)\)
\(=\left(a+5\right)\left(a^2+5a+6\right)=\left(a+5\right)\left(a^2+2a+3a+6\right)\)
\(=\left(a+5\right)\left(a+2\right)\left(a+3\right)\)
Và \(a\left(a+5\right)+4=a^2+5a+4=a^2+a+4a+4=\left(a+1\right)\left(a+4\right)\)
\(\Rightarrow A=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)=B\)
\(\Rightarrow P=\dfrac{A}{B}=1\)
tính tổng sau:
\(A=1\cdot2\cdot3\cdot.....\cdot2021-1\cdot2\cdot3\cdot.....\cdot2021-1\cdot2\cdot3\cdot.....\cdot2019\cdot2020^2\)
\(A=1.2.3.4...2019.\left(2020.2021-2020^2\right)=1.2.3.4...2019.2020\)
ĐỀ BÀI TOÁN LỚP 8
BÀI 1: Thu gọn đa thức sau:
F= \(\left(x-1\right)^3-x^2\left(x-3\right)\)
BÀI 2: Tìm x:
a) \(\left(x+3\right)^2=\left(x-2\right)\left(x+4\right)\)
b) \(\left(x+4\right)^2=2x^2+16\)
BÀI 3: Tính nhanh
\(A=2019\cdot2014-2016\cdot2017\)
\(B=2018^2-2016^2-4\cdot2016\)
\(C=2017\cdot2018-2016\cdot2019\)
Bài 1:
F=(x-1)3-x2(x-3)
=x3-3x2+3x-1-x3-3x2
=(x3-x3)-(3x2-3x2)+3x-1
=3x-1
Bài 2:
a)(x+3)2=(x-2)(x+4)
<=>x2+6x+9=x2+2x-8
<=>4x=-17
<=>x=-17/4
b)(x+4)2=2x2+16
<=>x2+8x+16=2x2+16
<=>8x=x2
<=>8x-x2=0
<=>x(8-x)=0
<=>x=0 hoặc x=8
Bài 1:
F=(x-1)3-x2(x-3)=x3-3x2+3x-1-x3+3x2=3x-1
Bài 2:
a, <=>(x+3)2-(x-2)(x-4)=0
<=>x^2+6x+9-x^2-4x+2x+8=0
<=>4x+17=0
<=>x=-4,25
b,<=>(x+4)2-2x2-16=0
<=>x2+8x+16-2x2-16=0
<=>8x-x2=0
<=>x(8-x)=0
<=>\(\orbr{\begin{cases}x=0\\x=8\end{cases}}\)
Bài 3:(đợi một xíu)
1 Tính giá trị của biểu thức
\(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\frac{1}{4}\right)^6\cdot2017}{\frac{1}{4096}\cdot\frac{1}{3}+\left(\frac{1}{2}\right)^{13}}\)=
1 Tính giá trị của biểu thức
\(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\frac{1}{4}\right)^6\cdot2017}{\frac{1}{4096}\cdot\frac{1}{3}+\left(\frac{1}{2}\right)^{13}}\)=
\(\frac{\left(\frac{1}{2}\right)^2.2018-\left(\frac{1}{4}\right)^2.2017}{\frac{1}{4096}.\frac{1}{3}+2^{13}}\)
=
\(\frac{\left(\frac{1}{2}\right)^2.2018-\left(\frac{1}{4}\right)^6.2017}{\frac{1}{4096}.\frac{1}{3}+2^{13}}\)\(\Leftrightarrow\frac{\left(\frac{1}{4}\right).2018-\left(\frac{1}{4096}\right).2017}{\frac{1}{4096}.\frac{1}{3}+2^{13}}\)
Lược bỏ các số giống nhau đi ta được :
\(\frac{\left(\frac{1}{4}\right).2018.2017}{\frac{1}{3}+2^{13}}\Leftrightarrow\frac{\left(\frac{1}{4}\right).2018.2017}{\frac{1}{3}.8192}\Leftrightarrow\frac{\frac{1}{4}.4070306}{\frac{8192}{3}}\)
\(=\frac{1017576,5}{\frac{8192}{3}}\)
1 Tính giá trị của biểu thức
\(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\frac{1}{4}\right)^6\cdot2017}{\frac{1}{4096}\cdot\frac{1}{3}-\left(\frac{1}{2}\right)^{13}}\)=
1 Tính giá trị của biểu thức
\(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\frac{1}{4}\right)^6\cdot2017}{\frac{1}{4096}\cdot\frac{1}{3}-\left(\frac{1}{2}\right)^{13}}\)=
= \(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\left(\frac{1}{2}\right)^2\right)^6\cdot2017}{\left(\frac{1}{2}\right)^2\cdot\frac{1}{3}\cdot\left(\frac{1}{2}\right)^{13}}\)
= \(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\frac{1}{2}\right)^{12}\cdot2017}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
=\(\frac{\left(\frac{1}{2}\right)^2\cdot\left(2018-2017\right)\cdot\left(\frac{1}{2}\right)^{10}}{\left(\frac{1}{2}\right)^{15}.\frac{1}{3}}\)
= \(\frac{\left(\frac{1}{2}\right)^2\cdot1\cdot\left(\frac{1}{2}\right)^{10}}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
= \(\frac{\left(\frac{1}{2}\right)^{12}}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
= \(\frac{1}{\left(\frac{1}{2}\right)^3\cdot\frac{1}{3}}\)
= \(\frac{1}{\frac{1}{24}}\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Tính giá trị của biểu thức sau:
\(D=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(D=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{2019.2021+1}{2019.2021}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2020.2020}{2019.2021}=\left(\dfrac{2}{1}.\dfrac{3}{2}...\dfrac{2020}{2019}\right).\left(\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}\right)=2020.\dfrac{2}{2021}=\dfrac{4040}{2021}\)
Câu 1. Tính hợp lý giá trị các biểu thức sau :
a. A = ( 689 - 31 ) - ( 269 - 131 )
b. B = \(\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}+1\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}\right)-\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}+1\right)\)c. C = \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!