\(\dfrac{5}{3}\)+\(\dfrac{1}{4}\)-\(\dfrac{8}{3}\) - \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)

(\(\dfrac{5}{3}-\dfrac{8}{3}\))+(\(\dfrac{1}{4}-\dfrac{7}{4}\)) + \(\dfrac{3}{2}\)

= -1 - \(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)

= -1

A = (\(x\) + 1)²⁰²² + (\(\sqrt{y-1}\))²⁰²³ đkxđ : y - 1 ≥ 0 ⇒ y ≥ 1

⇔ (\(x\) + 1)²⁰²² + (\(\sqrt{y-1}\))²⁰²³ = 0

vì (\(x\) + 1)²⁰²² ≥ 0; \(\sqrt{y-1}\) ≥ 0  ⇒ (\(\sqrt{y-1}\))²⁰²³ ≥ 0

Nên A = 0 ⇔ \(\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\)

             ⇔  \(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

Nghiệm của A là: \(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

Tìm GTNN chứ nhỉ e

\(D=\left|2022-x\right|+\left|x-1\right|\ge\left|2022-x+x-1\right|=2021\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2022-x\right)\left(x-1\right)\ge0\)

\(\Leftrightarrow1\le x\le2022\)

Vậy Min D=2021 \(\Leftrightarrow1\le x\le2022\)

Độ dài cạnh CD hay AB là: \(480:30=16\left(cm\right)\)

Độ dài CE là: \(30.2=60\left(cm\right)\)

Độ dài cạnh AD hay BC là: \(480:60=8\left(cm\right)\)

Chu vi hình bình hành ABCD là: \(\left(16+8\right).2=48\left(cm\right)\)

Cả mảnh vải dài :

  \(3,75\times100:75=5\left(m\right)\)

Nếu cắt đi 3/5 mảnh vải tức còn lại 2/5 mảnh vải . Vậy mảnh vải còn lại dài :

 \(5\times2:5=2\left(m\right)\)

`5x-5/6 =3x+2/3`

`=> 5x-3x=2/3+5/6`

`=>2x= 4/6 + 5/6`

`=> 2x= 9/6`

`=> x= 3/2 :2`

`=>x=3/2xx1/2`

`=>x= 3/4`

Vâỵ `x=3/4`

`@` `\text {Ans}`

`\downarrow`

`c)`

`5x - 5/6 = 3x + 2/3`

`=> 5x - 3x = 2/3 + 5/6`

`=> 2x = 3/2`

`=> x= 3/2 \div 2`

`=> x=3/4`

dấu ^ là gì vậy bạn?

là mũ ấy

\(\dfrac{2^{15}+2^{10}}{2^{10}+2^5}=\dfrac{2^{10}\left(2^5+1\right)}{2^5\left(2^5+1\right)}=\dfrac{2^{10}}{2^5}=2^5=32\)

 \(\dfrac{2^{15}+2^{10}}{2^{10}+2^5}\) = \(\dfrac{2^5\left(2^{10}+2^5\right)}{2^{10}+2^5}\) = 2⁵ = 32

\(-\dfrac{4}{3}\cdot x=\dfrac{2}{3}:\dfrac{7}{12}:\dfrac{4}{18}\)

\(\Rightarrow-\dfrac{4}{3}\cdot x=\dfrac{36}{7}\)

\(\Rightarrow x=\dfrac{\dfrac{36}{7}}{-\dfrac{4}{3}}=-\dfrac{27}{7}\)

-\(\dfrac{4}{3}\).\(x\) = \(\dfrac{2}{3}\): \(\dfrac{7}{12}\):\(\dfrac{4}{18}\)

-\(\dfrac{4}{3}.x\) = \(\dfrac{2}{3}\times\)\(\dfrac{12}{7}\)\(\times\)\(\dfrac{18}{4}\)

-\(\dfrac{4}{3}.\)\(x\)= \(\dfrac{36}{7}\)

     \(x\) = \(\dfrac{36}{7}\):(-\(\dfrac{4}{3}\))

    \(x\) = - \(\dfrac{27}{7}\)

Lời giải:

$C=\frac{5-x^2}{x^2+3}=\frac{8-(3+x^2)}{x^2+3}=\frac{8}{x^2+3}-1$

Ta thấy: $x^2\geq 0$ với mọi $x\in\mathbb{R}$ nên $x^2+3\geq 3$

$\Rightarrow \frac{8}{x^2+3}\leq \frac{8}{3}$

$\Rightarrow C=\frac{8}{x^2+3}-1\leq \frac{8}{3}-1=\frac{5}{3}$

Vậy $C_{\max}=\frac{5}{3}$ tại $x=0$