Tìm x biết: d) 25 x X = 9175
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a) \(A\left(x\right)=x^2-10x+25\)
\(\Rightarrow A\left(x\right)=\left(x-5\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}A\left(0\right)=\left(0-5\right)^2=25\\A\left(-1\right)=\left(-1-5\right)^2=36\end{matrix}\right.\)
b) \(A\left(x\right)+B\left(x\right)=6x^2-5x+25\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-A\left(x\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-\left(x^2-10x+25\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-x^2+10x-25\)
\(\Rightarrow B\left(x\right)=5x^2+5x\)
\(\Rightarrow B\left(x\right)=5x\left(x+1\right)\)
c) \(A\left(x\right)=\left(x-5\right)C\left(x\right)\)
\(\Rightarrow C\left(x\right)=\dfrac{\left(x-5\right)^2}{x-5}=x-5\left(x\ne5\right)\)
d) Nghiệm của B(x)
\(\Leftrightarrow B=0\)
\(\Leftrightarrow5x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) là nghiệm của B(x)
a ) 28 + x = − 59 x = − 59 − 28 x = − 87
b ) x − 4 x 2 − 25 = 0 ⇔ x − 4 = 0 x 2 − 25 = 0 ⇔ x = 4 x 2 = 25 ⇔ x = 4 x = ± 5
x + 2 = 17 ⇒ x + 2 = 17 x + 2 = − 17 ⇒ x = 15 x = − 19
d ) − 19 + 4 ( 2 − x ) = 25 − x − 19 + 8 − 4 x = 25 − x 4 x − x = − 19 + 8 − 25 3 x = − 36 = > x = − 12
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
a) 7x(x+1)−3(x+1)=0⇒(x+1)(7x−3)=07x(x+1)−3(x+1)=0⇒(x+1)(7x−3)=0
⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3
c) x2−10x=−25⇒x2−10x+
\(a.\dfrac{12+x}{42}=\dfrac{35}{42}\Leftrightarrow12+x=35\Leftrightarrow x=23\)
\(b.\dfrac{25-x}{40}=\dfrac{3}{8}\Leftrightarrow\dfrac{25-x}{40}=\dfrac{15}{40}\Leftrightarrow25-x=15\Leftrightarrow x=10\)
\(c.\dfrac{13+x}{20}=\dfrac{3}{4}\Leftrightarrow\dfrac{13+x}{20}=\dfrac{15}{20}\Leftrightarrow13+x=15\Leftrightarrow x=2\)
\(d.\dfrac{23-x}{25}=\dfrac{20}{25}\Leftrightarrow23-x=20\Leftrightarrow x=3\)
\(a,25\%x=42,6-3,28\\ \Leftrightarrow\dfrac{1}{4}x=39,32\\ \Leftrightarrow x=157,28\)
\(b,x:40\%=7,25+6,75\\ \Leftrightarrow x:40\%=14\\ \Leftrightarrow x=5,6\)
\(c,\dfrac{7011}{25}:\left(x-42,6\right)=117,14-47,03\\ \Leftrightarrow\dfrac{7011}{25}:\left(x-42,6\right)=70,11\\ \Leftrightarrow x-42,6=4\\ \Leftrightarrow x=46,6\)
\(d,0,3\left(x+\dfrac{1}{3}\right)=\dfrac{1}{2}\cdot\left(0,81:27\right)\\ \Leftrightarrow0,3\left(x+\dfrac{1}{3}\right)=0,015\\ \Leftrightarrow x+\dfrac{1}{3}=0,05\\ \Leftrightarrow x=-\dfrac{17}{60}\)
Lời giải:
$x:0,1+x:0,5-x:25\text{%}+x+x=2019$
$x\times 10+x\times 2-x\times 4+x+x=2019$
$x\times (10+2-4+1+1)=2019$
$x\times 10=2019$
$x=2019:10=201,9$
Đáp án A.