Tìm x thuộc Z, biết
a) 3x + 11 = x - 29
b) (x - 7)^2 = (-1)^ 2020
c) -34 < /x - 8/ < 2
d) (x + 5)^3 = -27
e) ( x + 3).(5x - 10)=0
Mk cần gấp ^_^
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\(a,\) Vì \(x,y\in Z\) nên \(\left(3x+2\right):3R2;R1\)
Mà \(\left(3x+2\right)\left(y-8\right)=12\) nên \(3x+2\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Do đó \(3x+2\in\left\{-4;-1;2\right\}\)
\(\Rightarrow x\in\left\{-2;-1;0\right\}\)
Với \(x=-2\Rightarrow\left(-4\right)\left(y-8\right)=12\Rightarrow y-8=-3\Rightarrow y=5\)
Với \(x=-1\Rightarrow\left(-3\right)\left(y-8\right)=12\Rightarrow y-8=-4\Rightarrow y=4\)
Với \(x=0\Rightarrow2\left(y-8\right)=12\Rightarrow y-8=6\Rightarrow y=14\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-2;5\right);\left(-1;4\right);\left(0;14\right)\)
\(b,\) Vì \(x,y\in Z\) nên \(\left(5x-4\right):5R1;R4\)
Mà \(\left(5x-4\right)\left(y+3\right)=-18\)
\(\Rightarrow5x-4\inƯ\left(-18\right)=\left\{-18;-9;-6;-3;-2;-1;1;2;3;6;9;18\right\}\\ \Rightarrow5x-4\in\left\{-9;1;6\right\}\\ \Rightarrow x\in\left\{-1;1;2\right\}\)
Với \(x=-1\Rightarrow-9\left(y+3\right)=-18\Rightarrow y+3=2\Rightarrow y=-1\)
Với \(x=1\Rightarrow y+3=18\Rightarrow y=15\)
Với \(x=2\Rightarrow6\left(y+3\right)=18\Rightarrow y+3=3\Rightarrow y=0\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-1;-1\right);\left(1;15\right);\left(2;0\right)\)
a: \(\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a) \(5\times\left(3+7\times x\right)=400\)
\(3+7\times x=80\)
\(7\times x=77\)
\(x=11\)
b) \(x\times37+x\times63=1200\)
\(x\times\left(37+63\right)=1200\)
\(x\times100=1200\)
\(x=12\)
c) \(x\times6+12:3=40\)
\(x\times6+4=40\)
\(x\times6=36\)
\(x=6\)
d) \(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=66\)
\(x+1=11\)
\(x=10\)
e) \(163:x+34:x=10\)
\(\left(163+34\right):x=10\)
\(197:x=10\)
\(x=19,7\)
a \(\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{1}{5}\\ \dfrac{2}{3}x=\dfrac{1}{5}-\dfrac{1}{3}\\ \dfrac{2}{3}x=\dfrac{-2}{15}\\ x=-\dfrac{2}{15}:\dfrac{2}{3}\\ x=-\dfrac{1}{5}\) b) \(\dfrac{4}{5}-\dfrac{5}{3}x=-2\\ \dfrac{5}{3}x=\dfrac{4}{5}+2\\ \dfrac{5}{3}x=\dfrac{14}{5}\\ x=\dfrac{14}{5}:\dfrac{5}{3}\\ x=\dfrac{42}{25}\)c) \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\\ \dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}\\ \dfrac{5}{3}:x=\dfrac{3}{10}\\ x=\dfrac{5}{3}:\dfrac{3}{10}\\ x=\dfrac{50}{9}\)d) \(\dfrac{5}{7}:x-3=-\dfrac{2}{7}\\ \dfrac{5}{7}:x=3-\dfrac{2}{7}\\ \dfrac{5}{7}:x=\dfrac{19}{7}\\ x=\dfrac{5}{7}:\dfrac{19}{7}\\ x=\dfrac{5}{19}\)
a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
a) 3x-x=-29-11
2x=-20
x=-20
b) (x-7)^2=1=1^2
x-7=1 hoặc x- 7 = -1
x=8 hoặc x=6
Vậy x thuộc {6;8}
c) vì /x-8/ lớn hơn hoặc bằng 0 với mọi x nên
TH1: /x-8/ = 0 suy ra x=8
TH2: /x-8/ = 1 suy ra x-8=1 hoặc x-8 = -1
x=9 hoặc x=7
Vậy x thuộc {7;8;9}
d) (x+5)^3=(-3)^3
x+5=-3
x=-8
vậy x=-8
e) (x+3).(5x-10)=0
x+3=0 hoặc 5x-10=0
x=-3 hoặc x=2
Vậy x thuộc {-3;2}