2.3 x 4 + 5 : 2 .5 =?
Giup mk nha!!!!!
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a, (x-2)(3x+5)=(2x-4)(x+1)
<=> (x-2)(3x+5)-2(x-2)(x+1)=0
<=>(x-2)(3x+5-2x-2)=0
<=>(x-2)(x+3)=0
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
(x-1)^2:5^21=25^30.5
(x-1)^2=25^30.5.5^21
=(5^2)^30.5^22
= 5^60.5^22
(x-1)^2 =5^82
(x-1)^2=(5^41)^2
x-1=5^41
x=5^41+1
2.3^x+1=10.3^12+8:3^12
2.3^x+1=10+8=18
3^x+1=18/2=9
3^x+1=3^2
x+1=2
x=1
cho bieu thuc A=\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}\)
hay chung to rang 5>A
giup mk nha
\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}\)
\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+...+\frac{5}{99}-\frac{5}{100}=\frac{5}{1}-\frac{5}{100}=\frac{99}{20}<\frac{100.}{20}=5\)
=>A<5 hay 5>A(đpcm)
Ta có
\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5.\left(1-\frac{1}{100}\right)\)
Vì \(1-\frac{1}{100}<1\)
=>\(5\left(1-\frac{1}{100}\right)<5.1=5\)
Vậy A<5
A = \(5.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
= \(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(5.\left(1-\frac{1}{100}\right)\)
= \(5-\frac{5}{100}<5\)
=> A < 5 (Đpcm).
Gọi biểu thức trên là \(A\). Ta có :
\(A=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=\frac{1}{4}+\frac{1}{2}-\frac{1}{6}\)
\(A=\frac{3}{4}-\frac{1}{6}\)
\(A=\frac{7}{12}.\)
Đáp số : \(\frac{7}{12}.\)
\(\frac{x}{3}+\frac{x}{4}=\frac{1}{5}\)
\(\Rightarrow\frac{20x}{60}+\frac{15x}{60}=\frac{12}{60}\)
\(\Rightarrow20x+15x=12\)
\(\Rightarrow35x=12\)
\(\Rightarrow x=\frac{12}{35}\)
\(5^4.125.\left(2,5\right)^{-5}.0,04\)
\(=5^4.5^3.\frac{32}{5^5}.\frac{1}{5^2}\)
\(=5^7.\frac{32}{5^7}=32\)
= 24 + 12,5
= 36,5
2.3 x 4 + 5 : 2.5
= 9.2 + 2
= 11.2