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a) \(N=x^2-10x+25\)
\(N=x^2-2\cdot5\cdot x+5^2\)
\(N=\left(x-5\right)^2\)
Thay x = 55 vào N ta có:
\(N=\left(55-5\right)^2=2500\)
b) \(P=\dfrac{x^4}{4}-x^2y+y^2\)
\(P=\left(\dfrac{x^2}{2}\right)^2-2\cdot\dfrac{x^2}{2}\cdot y+y^2\)
\(P=\left(\dfrac{x^2}{2}-y\right)^2\)
Thay x = 4 và \(y=\dfrac{1}{2}\) vào P ta có:
\(P=\left(\dfrac{4^2}{2}-\dfrac{1}{2}\right)^2=\dfrac{225}{4}\)
Phần b mình thấy kết quả nó sai b ạ thầy cho mình đáp án là 225/9
c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)
b: =>15-x=-10
hay x=25
a: =>-2x+17=9
=>-2x=-8
hay x=4
d: \(\Leftrightarrow9x^2=81\)
hay \(x\in\left\{3;-3\right\}\)
e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
`x^2=3`
`=>x=\sqrt{3}\or\x=-\sqrt{3}`
`x^2=36`
`<=>x^2=(+-6)^2`
`<=>x=+-6`
`x^2=25`
`<=>x^2=(+-5)^2`
`<=>x=+-5`
`2x^2+(-20)=55`
`<=>2x^2-20=55`
`<=>2x^2=75`
`<=>x^2=75/2`
`<=>x=+-\sqrt{75/2}`
`2(x-1)^2+5^0=9`
`<=>2(x-1)^2+1=9`
`<=>2(x-1)^2=8`
`<=>(x-1)^2=4`
`<=>x-1=2\or\x-1=-2`
`<=>x=3\or\x=-1`
a: =>-2x=9-17=-8
hay x=4
b: =>15-x=-10
hay x=25
d: \(\Leftrightarrow9x^2=81\)
hay \(x\in\left\{3;-3\right\}\)
e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
(87 + 3) : 3 = 30
25 + (42 – 11) > 55
100 < 888 : (4 + 4)
50 > (50 +50) : 5
`#qlv`
`(11)/(25) . (y + y5 - (23)/(55)) + 3/(14) . (56)/(25) = 1`
`=> (11)/(25) . (6y - (23)/(55)) + (12)/(25) = 1`
`=> (11)/(25) . (6y - (24)/(55)) = 1 - (12)/(25)`
`=> (11)/(25) (6y - (23)/(55)) = (13)/(25)`
`=> 6y - (23)/(55) = (13)/(25) : (11)/(25)`
`=> 6y - (23)/(55) = (13)/(11)`
`=> 6y = (13)/(11) + (23)/(55)`
`=> 6y = 8/5`
`=> y = 8/5 : 6`
`=> y = 8/5 . 1/6`
`=> y = 4/(15)`
Vậy `y = 4/(15)`
\(\dfrac{11}{25}\times\left(y+y\times5-\dfrac{23}{55}\right)+\dfrac{3}{14}\times\dfrac{56}{25}=1\)
\(\Rightarrow\dfrac{11}{25}\times\left(10\times y-\dfrac{23}{55}\right)+\dfrac{12}{25}-1=0\)
\(\Rightarrow\dfrac{22}{5}\times y-\dfrac{23}{125}-\dfrac{13}{25}=0\)
\(\Rightarrow\dfrac{22}{5}\times y=\dfrac{88}{125}\)
\(\Rightarrow y=\dfrac{88}{125}:\dfrac{22}{5}\)
\(\Rightarrow y=\dfrac{4}{25}\)
c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
*\(\left(45-25\right).\left(-11\right)+29.\left(-3-17\right)\)
\(=20.\left(-11\right)+29.\left(-20\right)\)
\(=-220+\left(-580\right)\)
\(=-800\)
*\(\left(-37\right).\left(55-23\right)-55.\left(23-37\right)\)
\(=-37.32-55.\left(-14\right)\)
\(=-1184-\left(-770\right)\)
\(=-1184+770\)
\(=-414\)
*\(\left(-12\right).\left(-125\right).\left(-4\right)\)
\(=1500.\left(-4\right)\)
\(=-6000\)
*\(\left(-1\right).2.\left(-3\right).4.\left(-5\right)\)
\(=-2.\left(-12\right).\left(-5\right)\)
\(=24.\left(-5\right)\)
\(=-120\)
`a: =(7/3-4/3)+11/5=16/5`
`b: =(18/13-5/13)+55/46=1+55/46=101/46`
`c: =27/25-4/9-2/25+5/9=(27/25-2/25)+(5/9-4/9)=1+1/9=10/9`