tìm x 4416 : x + 52 = 100
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xx +x+52=100
xx + x = 52 - 100
xx +x = 48
x *11 + x = 48
x *(11+1)=48
x * 12 = 48
x = 48 : 12
x= 4
b) /2x-5/=13
=>2x-5=13 hoặc -13
TH1: 2x-5=13 | TH2:2x-5=-13 |
2x=13+5=>2x=18 | 2x=-13+5=>2x=-8 |
x=18:2=>x=9 | x=-8:2=>x=4 |
a) \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(\Leftrightarrow30x+60-6x+30-24x=100\)
\(\Leftrightarrow\left(30x-6x-24x\right)+\left(60+30\right)\)
\(\Leftrightarrow x\left(30-6-24\right)+90=100\)
\(\Leftrightarrow0x+90=100\)
\(\Leftrightarrow0x=100-90\)
\(\Leftrightarrow0x=10\)
\(\Rightarrow\)Không có x thoả mãn.
b) \(|2x-5|=13\)
\(\Rightarrow\orbr{\begin{cases}2x-5=13\\2x-5=-13\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=18\\2x=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-4\end{cases}}\)
Vậy \(x\in\left\{-4;9\right\}\)
c) \(52-|x|=80\)
\(\Leftrightarrow|x|=52-80\)
\(\Leftrightarrow|x|=-28\)
Vì \(|x|\ge0\)
\(\Rightarrow\)Không có x thoả mãn.
2:
a: =-(x^2-12x-20)
=-(x^2-12x+36-56)
=-(x-6)^2+56<=56
Dấu = xảy ra khi x=6
b: =-(x^2+6x-7)
=-(x^2+6x+9-16)
=-(x+3)^2+16<=16
Dấu = xảy ra khi x=-3
c: =-(x^2-x-1)
=-(x^2-x+1/4-5/4)
=-(x-1/2)^2+5/4<=5/4
Dấu = xảy ra khi x=1/2
1)
a) \(A=x^2+4x+17\)
\(A=x^2+4x+4+13\)
\(A=\left(x+2\right)^2+13\)
Mà: \(\left(x+2\right)^2\ge0\) nên \(A=\left(x+2\right)^2+13\ge13\)
Dấu "=" xảy ra: \(\left(x+2\right)^2+13=13\Leftrightarrow x=-2\)
Vậy: \(A_{min}=13\) khi \(x=-2\)
b) \(B=x^2-8x+100\)
\(B=x^2-8x+16+84\)
\(B=\left(x-4\right)^2+84\)
Mà: \(\left(x-4\right)^2\ge0\) nên: \(A=\left(x-4\right)^2+84\ge84\)
Dấu "=" xảy ra: \(\left(x-4\right)^2+84=84\Leftrightarrow x=4\)
Vậy: \(B_{min}=84\) khi \(x=4\)
c) \(C=x^2+x+5\)
\(C=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}\)
\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\)
Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu "=" xảy ra: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=\dfrac{19}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(A_{min}=\dfrac{19}{4}\) khi \(x=-\dfrac{1}{2}\)
1:
a: A=x^2+4x+4+13
=(x+2)^2+13>=13
Dấu = xảy ra khi x=-2
b; =x^2-8x+16+84
=(x-4)^2+84>=84
Dấu = xảy ra khi x=4
c: =x^2+x+1/4+19/4
=(x+1/2)^2+19/4>=19/4
Dấu = xảy ra khi x=-1/2
ta có:\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{100}=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
bài toán được viết lại như sau:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right).x=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
\(\Rightarrow x=2012\left(\frac{1}{51}+...+\frac{1}{100}\right):\left(\frac{1}{51}+...+\frac{1}{100}\right)\)
\(\Rightarrow x=2012\)
vậy x=2012
= 23 x 0.07 + 0.07 x 52 + 0.07 x 25
= ( 23 + 52 + 25 ) x 0.07
= ( 75 + 25 ) x 0.07
= 100 x 0.07
= 7
23 x 0,07 + 7% x 52 + 7/100 x 25
= 23 x 0,07 + 0,07 x 52 + 0,07 x 25
= 0,07 x (23 + 52 + 25)
= 0,07 x 100
= 7
Chúc bạn học tốt.
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4416 : x + 52 = 100
4416 : x = 100 - 52
4416 : x = 48
x = 4416 : 48
x = 92
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