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a) \(10x+3920=29000\)
\(\Leftrightarrow10x=25080\)
\(\Leftrightarrow x=2508\)
b) \(\frac{x}{8}+2030=19202\)
\(\Leftrightarrow\frac{x}{8}=19202-2030\)
\(\Leftrightarrow x=\frac{4293}{2}\)
c) \(x^2+x^4=18\)
\(\Leftrightarrow x^2\left(1+x^2\right)=18\)
.... ( Nghiệm không được đẹp lắm, bài này mới lớp 6 thôi )
d) \(x+x+x+304=604\)
\(\Leftrightarrow3x=300\)
\(\Leftrightarrow x=100\)
e)\(x+5\frac{4}{5}=\frac{109}{5}\)
\(\Leftrightarrow x=16\)
a)\(\frac{5}{21}\)+\(\frac{-3}{7}\)<\(\frac{x}{21}\)<\(\frac{-2}{7}\)+\(\frac{8}{21}\)
\(\Rightarrow\)\(\frac{-4}{21}\)<\(\frac{x}{21}\)<\(\frac{2}{21}\)
\(\Rightarrow\)\(\frac{x}{21}\)\(\in\)\(\left\{\frac{-3}{21};\frac{-2}{21};\frac{-1}{21};\frac{0}{21};\frac{1}{21}\right\}\)
vậy x\(\in\)\(\left\{-3;-2;-1;0;1\right\}\)
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
Bài 6:
a: \(x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14-3}{21}=\dfrac{-17}{21}\)
d: \(x=\dfrac{9}{10}\cdot\dfrac{-5}{9}=\dfrac{-1}{2}\)
e: \(\Leftrightarrow x\cdot\dfrac{1}{3}=\dfrac{14}{21}-\dfrac{3}{21}=\dfrac{11}{21}\)
=>x=11/7
1. Tìm \(x\):
a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
\(\dfrac{x}{5}=\dfrac{1}{5}\)
\(\Rightarrow x=1\)
b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)
\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)
\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)
\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)
\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)
\(x=\dfrac{-17}{8}\)
c) \(2016^3.2016^x=2016^8\)
\(2016^x=2016^8:2016^3\)
\(2016^x=2016^{8-3}\)
\(2016^x=2016^5\)
\(\Rightarrow x=5\)
d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)
\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)
\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)
\(x+\dfrac{3}{4}=\dfrac{35}{4}\)
\(x=\dfrac{35}{4}-\dfrac{3}{4}\)
\(x=\dfrac{32}{4}=8\)
e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)
\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)
\(2,8.x-2^5=6\)
\(2,8.x=6+32\)
\(2,8.x=38\)
\(x=38:2,8\)
\(x=\dfrac{95}{7}\)
f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)
\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}:\dfrac{4}{7}\)
\(x=\dfrac{28}{15}\)
g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\)
\(x=\left(-6\right):3\)
\(x=-2\)
2. Thực hiện phép tính:
a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)
\(=\dfrac{7}{18}+\dfrac{9}{5}\)
\(=\dfrac{197}{90}\)
b) \(\dfrac{7.5^2-7^2}{7.24+21}\)
\(=\dfrac{7.25-7.7}{7.24+7.3}\)
\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)
\(=\dfrac{7.18}{7.27}\)
\(=\dfrac{2}{3}\)
c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
a) \(\left(-396\right)+2016+189+\left(-604\right)-109\)
\(=-396+2016+189-604-109\)
\(=-\left(396-2016\right)+189-604-109\)
\(=-\left(-1620\right)+189-604-109\)
\(=1620+189-604-109\)
\(=1809-604-109\)
\(=1205-109\)
\(=1096\)
b) \(10-\left|x+3\right|=-8-10\)
\(\Leftrightarrow10-\left|x+3\right|=-\left(8+10\right)\)
\(\Leftrightarrow10-\left|x+3\right|=-18\)
\(\Leftrightarrow\left|x+3\right|=10-\left(-18\right)\)
\(\Leftrightarrow\left|x+3\right|=10+18\)
\(\Leftrightarrow\left|x+3\right|=28\)
Xét trường hợp 1: \(x+3=28\)
\(\Rightarrow x=28-3\)
\(\Rightarrow x=25\)
Xét trường hợp 2: \(x+3=-28\)
\(\Rightarrow x=-28-3\)
\(\Rightarrow x=-\left(28+3\right)\)
\(\Rightarrow x=-31\)
Vậy \(x=25\) hoặc \(x=-31\)