Tìm x, biết :
130 - [ 4. ( x - 2 )2 + 15 ] = 79
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a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{5}\)
\(x=\dfrac{1}{2}-\dfrac{6}{5}\)
\(x=-\dfrac{7}{10}\)
b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)
\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)
\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)
\(x=\dfrac{60}{13}:\dfrac{12}{13}\)
\(x=5\)
Vì 15:5=3 mà X:5<15:5=3 nên:
X:5=0;1;2<3
=>X=0(Loại) vì (15-0).79>(15-2).79
=>X=5 hoặc X=10 thỏa mãn đ k đề bài đã cho.
Đ s :X=5 và X=10
79 - [25 + (x - 2)] = 15
=> 25 + (x - 2) = 79 - 15 = 64
=> x - 2 = 64 - 25 = 39
=> x = 39 + 2 = 41
a) x – 32 : 16 = 48 ó x – 2 = 48 ó x = 48 + 2 ó x = 50
b) 88 – 3.(7+x) = 64 ó 3.(7+x) = 88 – 64 ó 7 + x = 24:3 ó x = 8 – 7 ó x = 1
c) (5+4x) : 3 – 121 : 11 = 4 ó (5+4x) : 3 – 11 = 4 ó (5+4x) : 3 = 4 + 11 ó 5+4x = 15.3 ó 4x = 45 – 5 ó 4x = 40 ó x = 10
d) 15 – 2(3x+1) = 11.13 – 130 ó 15 – 2(3x+1) = 143 – 130 ó 15 – 2(3x+1) = 13
ó 2(3x+1) = 15 – 13 ó 3x + 1 = 2:2 ó 3x = 1 – 1 ó 3x = 0 ó x = 0
(\(\dfrac{3}{4}-2x\))4-2=79\(\rightarrow\left(\dfrac{3}{4}-2x\right)^4=79+2=81\)
\(\rightarrow\left(\dfrac{3}{4}-2x\right)^4=\left(\pm3\right)^4\)
\(\rightarrow\dfrac{3}{4}-2x=\pm3\)
-Khi \(\dfrac{3}{4}-2x=3\rightarrow2x=\dfrac{3}{4}-3=-\dfrac{9}{4}\rightarrow x=-\dfrac{9}{8}\)
-Khi \(\dfrac{3}{4}-2x=-3\rightarrow2x=\dfrac{3}{4}-\left(-3\right)=\dfrac{3}{4}+3=\dfrac{15}{4}\rightarrow x=\dfrac{15}{8}\)
\(\left(-4\right)^{x+3}-1=15\rightarrow\left(-4\right)^{x+3}=16=\left(-4\right)^2\)
\(\rightarrow x+3=2\rightarrow x=-1\)
\(125-\left(-75\right)+32-\left(48+32\right)\)
\(=125-\left(-75\right)+32-80\)
\(=125+75+32-80\)
\(=200+32-80\)
\(=232-80\)
\(=152\)
\(-63+415-37-115\)
\(=352-57-115\)
\(=315-115\)
\(=200\)
\(13.\left(-2\right)^2+37.4+4.\left(-150\right)\)
\(=13.2^2+37.4+4.-150\)
\(=13.4+37.4+4.-150\)
\(=4.\left(13+37+-150\right)\)
\(=4.-100\)
\(=-400\)
\(-15.x+4=79\)
\(\Leftrightarrow-15.x=79-4\)
\(\Leftrightarrow-15.x=75\)
\(\Leftrightarrow x=75:-15\)
\(\Leftrightarrow x=-5\)
Vậy \(x=-5\)
\(15-\left(2+x\right)=23\)
\(\Leftrightarrow2+x=15-23\)
\(\Leftrightarrow2+x=-8\)
\(\Leftrightarrow x=-8-2\)
\(\Leftrightarrow x=-10\)
Vậy \(x=-10\)
\(\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1^2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow x=\pm1\)
Vậy \(x\in\left\{\pm1\right\}\)
\(56:\left|x-4\right|=7\)
\(\Leftrightarrow\left|x-4\right|=56:7\)
\(\Leftrightarrow\left|x-4\right|=8\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=8\\x-4=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8+4\\x=-8+4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=12\\x=-4\end{cases}}}\)
Vậy \(x\in\left\{12;-4\right\}\)
a) 576 : x - 30 = 2
576 : x = 2 + 30
576 : x = 32
x = 576 : 32
x = 18
b) 180 - ( x - 45 ) : 2 = 120
( x - 45 ) : 2 = 180 - 120
( x - 45 ) : 2 = 60
x - 45 = 60 * 2
x - 45 = 120
x = 120 + 45
x = 165
a﴿ 576 : x ‐ 30 = 2
576 : x = 2 + 30
576 : x = 32
x = 576 : 32
x = 18
b﴿ 180 ‐ ﴾ x ‐ 45 ﴿ : 2 = 120
﴾ x ‐ 45 ﴿ : 2 = 180 ‐ 120
﴾ x ‐ 45 ﴿ : 2 = 60
x ‐ 45 = 60 * 2
x ‐ 45 = 120
x = 120 + 45
x = 165