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giá trị x>0 nguyên thỏa mãn: \(-\frac{7}{3}< \left|\frac{2}{7}-x\right|-\frac{5}{2}< -\frac{7}{4} \)
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
\(\Leftrightarrow\dfrac{1}{6}< \left|\dfrac{2}{7}-x\right|< \dfrac{3}{4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-\dfrac{2}{7}\right|>\dfrac{1}{6}\\\left|x-\dfrac{2}{7}\right|< \dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left(-\infty;\dfrac{10}{84}\right)\cup\left(\dfrac{38}{84};+\infty\right)\\x\in\left(-\dfrac{39}{84};\dfrac{87}{84}\right)\end{matrix}\right.\)
\(\Leftrightarrow x\in\left(\dfrac{38}{84};\dfrac{87}{84}\right)\)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
\(\frac{11}{14}+\left|\frac{2}{7}-x\right|-\frac{5}{2}=\frac{4}{3}\)
\(\Leftrightarrow\frac{11}{14}+\left|\frac{2}{7}-x\right|=\frac{23}{6}\)
\(\Leftrightarrow\left|\frac{2}{7}-x\right|=\frac{64}{21}\)
\(\Leftrightarrow\frac{2}{7}-x=\pm\frac{64}{21}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{2}{7}-x=\frac{64}{21}\\\frac{2}{7}-x=-\frac{64}{21}\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{58}{21}\\x=\frac{10}{3}\end{array}\right.\)
Mà \(x>0\)
Vậy \(x=\frac{10}{3}\)
Thanks!