Tìm x biết
7,5 * x + x * 3,5 = 39,93
x : 4+ x = 3,75
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7,5 * x + x * 3,5 = 39,93
x * (7,5 + 3,5) = 39,93
x * 11 = 39,93
x = 39,93 : 11
x = 3,63
x : 4 + x = 3,75
x : (4 + 1) = 3,75
x : 5 = 3,75
x = 3,75 * 5
x = 18,75
(7,5+3,5)*x=39,93 x:4+x=3,75
11*x=39,93 x:4+x:1=3,75
x=39,93:11 x:(4+1)=3,75
x=3,63 x:5=3,75 x=3,75*5 x= 18,75
\(a.7,5\times x+x\times3,5=39,93\)
\(\Rightarrow x\times\left(7,5+3,5\right)=39,93\)
\(\Rightarrow x\times11=39,93\)
\(\Rightarrow x=39,93:11\)
\(\Rightarrow x=3,63\)
\(b.x:4+x=3,75\)
\(\Rightarrow x:\frac{4}{1}+x=3,75\)
\(\Rightarrow x\times\frac{1}{4}+x=3,75\)
\(\Rightarrow x\times\left(\frac{1}{4}+1\right)=3,75\)
\(\Rightarrow x\times1,25=3,75\)
\(\Rightarrow x=3,75:1,25\)
\(\Rightarrow x=3\)
a) 7,5 *x + x*3,5=39,93
(7,5 + 3,5) * x=39,93
11 * x = 39,93
x= 39,93 : 11
x= 3,63
b)x : 4 + x=3,75
x * 1/4 + x *1 =3,75
x * (1/4 + 1) = 3,75
x * 5/4 = 3,75
x = 3,75 : 5/4
x= 3
= ( 7,5 + 3,5 ) * x = 39,93
= 11 * x = 39,93
x = 39,93 : 11
x = 3,69
x : 4 + x = 3,75
x : ( 4 + 1 ) = 3,75
x : 5 = 3,75
x = 3,75 x 5
x = 18,75
\(a,\frac{x+15}{x}=\frac{4}{3}\Rightarrow4x=3x+45\Leftrightarrow x=45\)
\(b,\frac{7,5-x}{3,5+x}=\frac{5}{6}\Rightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Rightarrow x=2,5\)
\(c,\frac{x-20}{x-10}=\frac{x+40}{x+70}\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)
\(\Leftrightarrow x^2+50x-1400=x^2+30x-400\)
\(\Leftrightarrow20x=1000\)
\(\Rightarrow x=50\)
a. \(\frac{\left(x+15\right)}{x}=\frac{4}{3}\Leftrightarrow4x=3\left(x+15\right)\Leftrightarrow4x=3x+45\Leftrightarrow x=45\)
Vậy x=45
b. \(\frac{7,5-x}{3,5+x}=\frac{5}{6}\Leftrightarrow5\left(3,5+x\right)=6\left(7,5-x\right)\Leftrightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Leftrightarrow x=2,5\)
Vậy x=2,5
c. \(\frac{x+20}{x-10}=\frac{x+40}{x+70}\Leftrightarrow\left(x+40\right)\left(x-10\right)=\left(x+20\right)\left(x+70\right)\)
\(\Leftrightarrow x^2+30x-400=x^2+90x+1400\Leftrightarrow-60x=-30\Leftrightarrow x=-30\)
Vậy x=-30
Ta có : \(\frac{\left(11,2-10-1,2\right)x\left(3,75-0,75\right)}{2011+3014}=\frac{0x3}{2011+3014}=\frac{0}{2011+3014}=0\)
\(\Rightarrow\left(2000\times7,5+2012:3\right)\times\left(21-3,5\times0,25\right)\times\frac{\left(11,2-10-1,2\right)\times\left(3,75-0,75\right)}{2011+3014}\)
\(=\left(2000\times7,5+2012:3\right)\times\left(21-3,5\times0,25\right)\times0\)
\(=0\)
a) Đề sai
b) Vì |x + 2,8| \(\ge\)0 \(\forall\)x
=> |x + 2,8| - 3,5 \(\ge\)-3,5 \(\forall\)x
Dấu " = " xảy ra khi và chỉ khi |x + 2,8| = 0 => x = -2,8
Vậy Bmin = -3,5 khi x = -2,8
1). 7,5 x X + X x 3,5 = 39,93
X x (7,5 + 3,5) = 39,93
X x 11 = 39,93
X = 39,93 : 11
X = 3,63
2) x : 4 + x = 3,75
x : 4 + x : 1 = 3,75
x : (4 + 1) = 3,75
x : 5 = 3,75
x = 3,75 : 5 = 0,75
x ; 4 + x = 3,75
x * 0,25 + x = 3,75
x * (0,25 + 1) = 3,75
x * 1,25 = 3,75
x = 3,75/1,25
x = 3