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1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
a) \(6-2\left(x+4\right)=-4\)
\(\Leftrightarrow6-2x-8=-4\)
\(\Leftrightarrow-2-2x=-4\)
\(\Leftrightarrow-2x=-4+2\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b) tự làm.
c) \(2x-15:5=-7\)
\(\Leftrightarrow2x-7,5=-7\)
\(\Leftrightarrow2x=-7+7,5\)
\(\Leftrightarrow2x=0,5\)
\(\Leftrightarrow x=0,25\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4}\)
d) \(231-\left(x-6\right)=1339:13\)
\(\Leftrightarrow231-x+6=103\)
\(\Leftrightarrow237-x=103\)
\(\Leftrightarrow237-x+6=103\)
\(\Leftrightarrow-x=103-237\)
\(\Leftrightarrow-x=-134\)
\(\Leftrightarrow x=134\)
e) đề j mà có 2 dấu bằng >> sai :v
f) \(28-x^3=1\)
\(\Leftrightarrow-x^3=1-28\)
\(\Leftrightarrow-x^3=-27\)
\(\Leftrightarrow x^3=27\)
\(\Leftrightarrow x^3=3^3\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
g) \(\left(x+2\right)^3=27\)
\(\Leftrightarrow\left(x+2\right)^3=3^3\)
\(\Leftrightarrow x+2=3\)
\(\Leftrightarrow x=3-2\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
h) \(81-\left(x-5\right)^2=56\)
\(\Leftrightarrow-\left(x-5\right)^2=56-81\)
\(\Leftrightarrow-\left(x-5\right)^2=-25\)
\(\Leftrightarrow\left(x-5\right)^2=25\)
\(\Leftrightarrow x-5=\pm5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=5\\x-5=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5+5\\x=-5+5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=0\end{matrix}\right.\)
Vậy \(x_1=0;x_2=10\)
j) \(x^2=9\)
\(\Leftrightarrow x=\pm3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(x_1=-3;x_2=3\)
l) tự làm.
a) \(6-2x-8=-4\Leftrightarrow2x=2\Leftrightarrow x=1\)
b) \(6⋮x\Rightarrow x=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
c) \(2x-3=-7\Leftrightarrow2x=-4\Leftrightarrow x=-2\)
d) \(231-x+6=103\Leftrightarrow x=124\)
e) hình như sai đề
f) \(x^3=27\Leftrightarrow x=3\)
g) \(x+2=3\Leftrightarrow x=1\)
h) \(\left(x+5\right)^2=25\Leftrightarrow x+5=5\Leftrightarrow x=0\)
i) \(x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)
j) \(14⋮\left(x+3\right)\Rightarrow x=\left\{-17;-10;-5;-4;-2;4;11\right\}\)