12 - ( x + 5) = 7
Giúp mình vs
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a)x\(\in\left\{-5;-4;-3;-2;-1\right\}\)
b)x\(\in\left\{-2;-1;0;1;2;3\right\}\)
c)x\(\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)
d)x\(\in\left\{1;2;3;4\right\}\)
e)x\(\in\left\{0;1;2;3;4\right\}\)
f)x\(\in\left\{-10;-9;-8;-7\right\}\)
=>|3x-2/5|=1/35+90/35=91/35
=>3x-2/5=91/35 hoặc 3x-2/5=-91/35
=>3x-2/5=13/5 hoặc 3x-2/5=-13/5
=>3x=15/5=3 hoặc 3x=-11/5
=>x=-11/5 hoặc x=1
Lời giải:
$|3x-\frac{2}{5}|=\frac{1}{35}+\frac{18}{7}=\frac{13}{5}$
$\Rightarrow 3x-\frac{2}{5}=\frac{13}{5}$ hoặc $3x-\frac{2}{5}=\frac{-13}{5}$
$\Rightarrow 3x=3$ hoặc $3x=\frac{-11}{5}$
$\Rightarrow x=1$ hoặc $x=\frac{-11}{15}$
\(a.\dfrac{6}{7}-\dfrac{4}{21}=\dfrac{18-4}{21}=\dfrac{14}{21}=\dfrac{2}{3}\\ b.6\times\dfrac{12}{18}=\dfrac{6\times12}{18}=\dfrac{4}{1}=4\\ c.\dfrac{2}{5}\times\dfrac{3}{7}+\dfrac{2}{5}\times\dfrac{4}{7}-\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\times\dfrac{2}{5}=\dfrac{7}{7}\times\dfrac{2}{5}=1\times\dfrac{2}{5}=\dfrac{2}{5}\)
a: \(x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)
Tổng là 0
b: \(x\in\left\{-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7\right\}\)
Tổng là 7
\(=\dfrac{3}{7}\cdot\left(\dfrac{4}{9}+\dfrac{5}{9}+1\right)=\dfrac{3}{7}\cdot2=\dfrac{6}{7}\)
\(=\dfrac{3}{7}\times\dfrac{4}{9}\times\dfrac{5}{9}\times\dfrac{3}{7}+\dfrac{3}{7}=\dfrac{3}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}+1\right)=\dfrac{3}{7}\times2=\dfrac{6}{7}\)
\(0,3:2,5=3:25\)
\(4\dfrac{2}{5}:1\dfrac{1}{3}=\dfrac{22}{5}:\dfrac{4}{3}=33:10\)
\(-3,2:1\dfrac{2}{7}=\dfrac{-16}{5}:\dfrac{9}{7}=112:45\)
\(A=-x\left(x-6\right)+7\)
\(=-x^2+6x+7\)
\(=-\left(x^2-6x-7\right)\)
\(=-\left(x^2-6x+9-16\right)\)
\(=-\left(x-3\right)^2+16\le16\forall x\)
Dấu '=' xảy ra khi x=3
x+ 5=12-7
x+5=5
x=0
\(x+5=5\Leftrightarrow x=0\)