1/2×x+2/3×x−1=3 1/3
giup mik luon voi a
1/2*x^2.(-2*x^2*y^2*z),(-1/3)*x^2*y^3
giup mk voi a
`1/2x^2(-2x^2y^2z).(-1/3)x^2y^3`
`=1/2 .(-2).(-1/3)x^{2+2+2}.y^{2+3}.z`
`=1/3x^6y^5z`
Ta có: \(\dfrac{1}{2}x^2\cdot\left(-2x^2y^2z\right)\cdot\left(-\dfrac{1}{3}\right)x^2y^3\)
\(=\left(\dfrac{1}{2}\cdot2\cdot\dfrac{1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)
\(=\dfrac{1}{3}x^6y^5z\)
bai 5 : tinh
a) tim x , biet (x+1) +(x+2 ) + ...+(x+100)=5750
b) chung minh rang B = 1/2^2 + 1/3^2 + 1/4^2 + ...+1/2021^2 < 1
giup mik luon voi
\(∘backwin\)
\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)
\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)
\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)
\(100 x + 5050 = 5750\)
\( 100 x = 5750 − 5050\)
\(100 x = 700\)
\(x = 700 : 100\)
\(x = 7\)
\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)
\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(B<1-\)\(\dfrac{1}{2021}\)
\(B<\)\(\dfrac{2020}{2021}\)
\(\dfrac{2020}{2021}< 1\)
\(B<1\)
a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7
b) Ta thấy: 1/2^2 < 1/2.3
1/3^2 < 1/3.4
...
1/2021^2 < 1/2021.2022
--> B=1/2^2 + 1/3^2 + 1/4^2 + ...+ 1/2021^2 < 1/2.3 + 1/3.4 + ... +1/2021.2022 (1)
Ta có: 1/2.3 + 1/3.4 + ... +1/2021.2022
=1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2021 - 1/2022
=1/2 - 1/2022 < 1 (2)
Từ (1) và (2) --> B<1 (đpcm)
<
x-2/4 = x -1/3
giup minh voi
11/8−3/8×x=1/8
giup mik luon voi
\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:\dfrac{3}{8}\)
\(x=\dfrac{10}{3}\)
\(\dfrac{3}{8}x=\dfrac{11}{8}-\dfrac{1}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:\dfrac{3}{8}=\dfrac{5}{4}.\dfrac{8}{3}=\dfrac{10}{3}\)
` 11/8 - 3/8 . x = 1/8 <=> 2/8 . x = 11/8 - 1/8 <=> 2/8x = 10/8 <=> x = 10/8 : 2/8 <=> x = 5 `
5/7 x X + 2/5 = 2/3
Giup em voi a
\(\dfrac{5}{7}\times x+\dfrac{2}{5}=\dfrac{2}{3}\\ \dfrac{5}{7}\times x=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10}{15}-\dfrac{6}{15}=\dfrac{4}{15}\\ x=\dfrac{4}{15}:\dfrac{5}{7}=\dfrac{4\times7}{15\times5}=\dfrac{28}{75}\)
Vậy `x=28/75`
`HaNa☘D`
\(\dfrac{5}{7}\times x+\dfrac{2}{5}=\dfrac{2}{3}\)
\(\dfrac{5}{7}\times x=\dfrac{2}{3}-\dfrac{2}{5}\)
\(\dfrac{5}{7}\times x=\dfrac{10}{15}-\dfrac{6}{15}\)
\(\dfrac{5}{7}\times x=\dfrac{4}{15}\)
\(x=\dfrac{4}{15}:\dfrac{5}{7}\)
\(x=\dfrac{4}{15}\times\dfrac{7}{5}\)
\(x=\dfrac{28}{75}\)
#Toru
=>x*5/7=2/3-2/5=4/15
=>x=4/15:5/7=4/15*7/5=28/75
cho don thuc 3(a+1/a)x^4y^2 voi a la hang so ,a khac 0
a)tim a de da thuc luon khong am voi moi x,y
b)tim a de da thuc luon khong duong voi moi x,y
1) Phan tich da thuc sau thanh nhan tu: x2-x-2008.2009
2) Chung minh rang voi moi x,y,z ta luon co: x2+4y2+z2>=2x+12y+4z
3) Cho a-b=4. Tinh gia tri cua bieu thuc: a3-12ab-b3
cac ban lam duoc cau nao thi giup mik nha. mik dang can gap lam
cho ham so f(x) xac dinh voi x thuoc R biet voi moi x ta luon co f(x) +3 x f(1/x) =x^2. tinh f(x)
Chung minh bieu thuc Q=(x^4*y^n+1-1/2*x^3*y^n+2):1/2x^3*y^n-20x^4*y:5*xy^2 (n thuoc N) luon <0 voi moi gia tri x khac 0,y khac 0