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Bài 3: Tìm x, biết:
a) \(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)
\(\Leftrightarrow40x-40=0\)
\(\Leftrightarrow4x=40\)
\(\Leftrightarrow x=10\)
Vậy x = 10
b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(\Leftrightarrow\left(2x+3\right)^2-4\left(x^2-1\right)=49\)
\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)
\(\Leftrightarrow12x-36=0\)
\(\Leftrightarrow12x=36\)
\(\Leftrightarrow x=3\)
Vậy x = 3
c) \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18\)
\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=18\)
\(\Leftrightarrow2\left(1-2x\right)=18\)
\(\Leftrightarrow2-4x=18\)
\(\Leftrightarrow4x=-16\)
\(\Leftrightarrow x=-4\)
Vậy x =-4
d) \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)
\(\Leftrightarrow12x-5=0\)
\(\Leftrightarrow12x=5\)
\(\Leftrightarrow x=\frac{5}{12}\)
Vậy \(x=\frac{5}{12}\)
e) \(\left(x-5\right)^2-x\left(x-4\right)=9\)
\(\Leftrightarrow x^2-10x+25-x^2+4x=9\)
\(\Leftrightarrow25-6x=9\)
\(\Leftrightarrow6x=16\)
\(\Leftrightarrow x=\frac{8}{3}\)
Vậy \(x=\frac{8}{3}\)
f) \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)
\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)
\(\Leftrightarrow21-5x=0\)
\(\Leftrightarrow5x=21\)
\(\Leftrightarrow x=\frac{21}{5}\)
Vậy \(x=\frac{21}{5}\)
a: \(\left|\dfrac{1}{2}x-3\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=\dfrac{1}{2}\\\dfrac{1}{2}x-3=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{7}{2}\\\dfrac{1}{2}x=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
b: \(\left|3x+\dfrac{1}{5}\right|-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Leftrightarrow\left|3x+\dfrac{1}{5}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{1}{5}=1\\3x+\dfrac{1}{5}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{4}{5}\\3x=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{15}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: \(\left|\dfrac{1}{2}x-3\right|< \dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-3>-\dfrac{1}{2}\\\dfrac{1}{2}x-3< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x>\dfrac{5}{2}\\\dfrac{1}{2}x< \dfrac{7}{2}\end{matrix}\right.\Leftrightarrow5< x< 7\)
e: \(\left|3x-\dfrac{4}{5}\right|>\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}>\dfrac{1}{3}\\3x-\dfrac{4}{5}< -\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x>\dfrac{17}{15}\\3x< \dfrac{7}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{17}{45}\\x< \dfrac{7}{45}\end{matrix}\right.\)
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
a: \(\Leftrightarrow\left(x-1\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
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 ) 234 + 450 = x + 4
=> 684 = x + 4
=> x + 4 = 684
=> x = 680
b ) 208 > x + 65 > 206
=> x + 65 = 207
=> x = 142
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