Question

розв'язати рівняння 2, 3 , 4

Answer 1

Ответ:

$90;$

$22;$

$1;$

Объяснение:

$((4x+174) \cdot 15+90):45=180;$

$(4x+174) \cdot 15+90=180 \cdot 45;$

$(4x+174) \cdot 15=180 \cdot 45-90 \quad | \quad :15$

$4x+174=180 \cdot \dfrac{45}{15}-\dfrac{90}{15};$

$4x+174=180 \cdot 3-6;$

$4x=540-6-174;$

$4x=540-180;$

$4x=360;$

$x=90;$

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$200: \bigg (\bigg ( \bigg (7+\dfrac{154}{x} \bigg )+132 \bigg ) \cdot 5-630 \bigg )=2;$

$\bigg ( \bigg (7+\dfrac{154}{x} \bigg )+132 \bigg ) \cdot 5-630=200:2;$

$\bigg ( \bigg (7+\dfrac{154}{x} \bigg )+132 \bigg ) \cdot 5-630=100;$

$\bigg ( \bigg (7+\dfrac{154}{x} \bigg )+132 \bigg ) \cdot 5=630+100;$

$\bigg ( \bigg (7+\dfrac{154}{x} \bigg )+132 \bigg ) \cdot 5=730;$

$\bigg (7+\dfrac{154}{x} \bigg )+132=730:5;$

$7+\dfrac{154}{x}+132=146;$

$\dfrac{154}{x}=146-132-7;$

$\dfrac{154}{x}=7;$

$x=\dfrac{154}{7};$

$x=22;$

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$\bigg (\bigg (\dfrac{(120+x) \cdot 40}{2}+200 \bigg ):131 \bigg ):20=1;$

$((120+x) \cdot 20+200):131=1 \cdot 20;$

$(120+x) \cdot 20+200=20 \cdot 131;$

$(120+x) \cdot 20=20 \cdot 131-20 \cdot 10;$

$(120+x) \cdot 20=20 \cdot (131-10);$

$120+x=121;$

$x=1;$

Answer 2

на фото................