10³ + 2¹⁵

= 1000 + 32768

= 33768

Mà 33768 : 33 = 1023 (dư 9)

Em xem lại đề

\(2^{x+1}-2^x=3^2\)

\(\Rightarrow2^x\cdot\left(2-1\right)=9\)

\(\Rightarrow2^x=9\)

\(\Rightarrow x\in\varnothing\)

\(2^{x+1}-2^x=3^2\)

=>2^x*2-2^x=9

=>2^x=9

=>\(x\in\varnothing\)

a: \(=25x^4-10x^3+5x^2\)

c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)

A = |\(x\) + 5| + 2023

|\(x\) + 5| ≥ 0 ⇒| \(x\) + 5| + 2023 ≥ 2023⇒ A(min) = 2023 xảy ra khi \(x\) = -5

B = (\(x+2\))² - 2023

(\(x\) + 2)2 ≥ 0 ⇒ (\(x\) + 2)2 ≥ - 2023 ⇒ A(min) = -2023  xảy ra khi \(x\) = -2

C = \(x^2\) - 6\(x\) + 20

C = (\(x^2\) - 3\(x\)) - ( 3\(x\) - 9) + 11

C = \(x\)(\(x-3\)) - 3(\(x\) -3) + 11

C = (\(x-3\))(\(x\)-3) + 11

C = (\(x-3\))² + 11

(\(x\) -3)² ≥ 0 ⇒ (\(x\) - 3)² + 11 ≥ 11 vậy C(min) = 11 xảy ra khi \(x=3\)

D = \(x^2\) + 10\(x\) - 25

D = \(x^2\) + 5\(x\) + 5\(x\) + 25 - 55

D = (\(x^2\) + 5\(x\)) + (5\(x\) + 25) - 50

D = \(x\)(\(x\) + 5) + 5(\(x\) + 5)  - 50

D = (\(x\) +5)(\(x\) + 5) - 50

D = ( \(x\) + 5)² - 50

(\(x+5\))² ≥ 0 ⇒ (\(x\) + 5)² - 50 ≥ -50 ⇒ D(min) = -50 xảy ra khi \(x\) = -5

\(2^2\cdot3\left(x+5\right)-6^2=\left(2^3+2^2\right)\cdot2^2\)

\(\Rightarrow4\cdot\left(3x+15\right)-36=12\cdot4\)

\(\Rightarrow12x+60-36=48\)

\(\Rightarrow12x+24=48\)

\(\Rightarrow12x=24\)

\(\Rightarrow x=24:12=2\)

22⋅3(x+5)−62=(23+22)⋅22

$\Rightarrow 4\cdot \left(3x+15\right)-36=12\cdot 4$

$\Rightarrow 12x+60-36=48$

$\Rightarrow 12x+24=48$

$\Rightarrow 12x=24$

$\Rightarrow x=24:12=2$

\(x^2+x=0\)\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy \(x=-1\)hoặc \(x=0\)

\(x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

`@` `\text {Ans}`

`\downarrow`

`(2^2+1) \times (x+14) = 5^2 \times 4 + (2^5 + 3^2 + 7^2) \div 2`

` \Rightarrow (4+1) \times (x+14) = 5^2\times 2^2 + ( 32 + 9 + 49) \div 2`

`\Rightarrow 5 \times (x+14) = (5*2)^2 + (32+58) \div 2`

`\Rightarrow 5 \times (x+14) = 10^2+90 \div 2`

`\Rightarrow 5 \times (x+14) = 100 + 45`

`\Rightarrow 5 \times (x+14) = 145`

`\Rightarrow x+14 = 145 \div 5`

`\Rightarrow x+14=29`

`\Rightarrow x=29-14`

`\Rightarrow x=15`

Vậy, `x=15.`

\(\left(2^2+1\right)\cdot\left(x+14\right)=5^2\cdot4+\left(2^5+3^2+7^2\right):2\)

\(\Rightarrow\left(4+1\right)\cdot\left(x+14\right)=25\cdot4+\left(32+9+49\right):2\)

\(\Rightarrow5\cdot\left(x+14\right)=100+45\)

\(\Rightarrow5x+70=145\)

\(\Rightarrow5x=75\)

\(\Rightarrow x=\dfrac{75}{5}=15\)