(x+1)+(x+2)+(x+3)+...........+(x+10)=165
giúp mik với
thank you nha
yêu nhiều nhoa nhoa
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK : x≥0
Ta có A=\(\frac{3-\sqrt{x}}{1+\sqrt{x}}\)
=\(\frac{-\left(\sqrt{x}+1\right)+4}{\sqrt{x}+1}\)
=\(-1+\frac{4}{\sqrt{x}+1}\)
Ta có x ≥ 0
⇒\(\sqrt{x}\) ≥ 0
⇒\(\sqrt{x}\) + 1 ≥ 1
⇒\(\frac{1}{\sqrt{x}+1}\) ≤ \(\frac{1}{1}\)
⇒\(\frac{4}{\sqrt{x}+1}\) ≤ 4
⇒-1 + \(\frac{4}{\sqrt{x}+1}\) ≤ -1 + 4 = 3
⇒ A ≤ 3
Dấu "=" xảy ra khi : x = 0
Vậy Amax=3 khi x = 0
1. 89:x-51,5:x=62,5
( 89-51,5 ):x=62,5
37,5:x=62,5
x=62,5.37,5
x=1343,75
2.34,56 x 999 + 3,456 x 10
=34,56x999+34,56x1
=34,56x(999+1)
=34,56x1000
=34560
1.
89:x-51.5:x=62.5
89.1/x-51,5.1/x=62,5
1/x(89-51,5)=62,5
1/x.37,5=62,5
1/x=5/3
x=3/5
2 tương tự nhân vô
a) \(\left|x-9\right|=2x+5\)
khi \(x\ge-\frac{5}{2}\), ta có:
\(\orbr{\begin{cases}x-9=2x+5\\x-9=-2x-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=14\\3x=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=-14\\x=\frac{4}{3}\end{cases}}\)
\(x=-14\)không thỏa mãn
\(x=\frac{4}{3}\)thỏa mãn
vậy x=4/3
1) \(3^x=\dfrac{9^8}{27^3\cdot81^2}\)
\(\Rightarrow3^x=\dfrac{\left(3^2\right)^8}{\left(3^3\right)^3\cdot\left(3^4\right)^2}\)
\(\Rightarrow3^x=\dfrac{3^{16}}{3^{15}}\)
\(\Rightarrow3^x=3\)
\(\Rightarrow x=1\)
2) \(\dfrac{2^{4-x}}{16^5}=32^6\)
\(\Rightarrow\dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
\(\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\)
\(\Rightarrow2^{4-x}=2^{20}\cdot2^{30}\)
\(\Rightarrow2^{4-x}=2^{50}\)
\(\Rightarrow4-x=50\)
\(\Rightarrow x=-46\)
3) \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)
\(\Rightarrow\dfrac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)
\(\Rightarrow\dfrac{2^{2x-3}}{2^{20}}=2^{29}\)
\(\Rightarrow2^{2x-3}=2^{49}\)
\(\Rightarrow2x-3=49\)
\(\Rightarrow2x=52\)
\(\Rightarrow x=26\)
\(1;\left|x\right|-x=0\)
\(\Leftrightarrow\left|x\right|=x\Rightarrow x=\pm x\)
\(\Leftrightarrow\orbr{\begin{cases}x=x\\x=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x\inℝ\\x=0\end{cases}}}\)
\(2;\left|x\right|-x=2\)
\(\Leftrightarrow\left|x\right|=2+x\Leftrightarrow x=\pm\left(x+2\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=x+2\\x=-x-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x\in\varnothing\\x+x=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\in\varnothing\\2x=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\in\varnothing\\x=1\end{cases}}}\)
3) Nếu \(x\ge0\)thì \(pt\Leftrightarrow3x+x=16\)
\(\Leftrightarrow4x=16\Leftrightarrow x=4\)
Nếu \(x< 0\)thì \(pt\Leftrightarrow-3x+x=16\)
\(\Leftrightarrow-2x=16\Leftrightarrow x=-8\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+10\right)=165.\)
\(\Leftrightarrow10x+55=165\)
\(\Leftrightarrow10x=110\Leftrightarrow x=11\)
=> 10.x + (1+2+3+4+5+6+7+8+9+10)=165
=>10.x +55=165
=> 10.x =165-55=110
=> x=110:10
=> x=11
Mk làm hơi tắt mong bạn thông cảm