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NN
4
1 tháng 2 2019
CÁCH KHÁC:
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)
\(<=>x\left(x+10\right)\left(x+4\right)\left(x+6\right)+128\)
\(<=>\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)
\(<=>\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128\)
\(<=>\left(x^2+10x\right)^2+2.\left(x^2+10x\right).12+12^2-16\)
\(<=>\left(x^2+10x+12\right)^2-4^2\)
\(<=>\left(x^2+10x+12-4\right) \left(x^2+10x +12+4\right)\)
\(<=>\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)
\(<=>\left(x^2+10x+8\right)\left(x^2+2x+8x+16\right)\)
\(<=>\left(x^2+10x+8\right)\left[x\left(x+2\right)+8\left(x+2\right)\right]\)
\(<=>\left(x^2+10x+8\right)\left(x+2\right)\left(x+8\right)\)
\(< =>\left[{}\begin{matrix}x^2+10x+8=0\\x+2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-5+\sqrt{17}\\x=-5-\sqrt{17}\\x=-2\\x=-8\end{matrix}\right.\)
Vậy...
KB
1 tháng 2 2019
Ta có :
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=0\)
\(\Leftrightarrow\left[x\left(x+10\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)\left(x^2+10x+24\right)+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)^2+24\left(x^2+10x\right)+144=16\)
\(\Leftrightarrow\left(x^2+10x+12\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x+12=4\\x^2+10x+12=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2-13=4\\\left(x+5\right)^2-13=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=17\\\left(x+5\right)^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+5=\pm\sqrt{17}\\x+5=\pm3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{17}-5\\\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\end{matrix}\right.\)
VD
18 tháng 3 2022
\(a,2x-5=-x+4\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\\ b,\left(4x-10\right)\left(25+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\25+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-5\end{matrix}\right.\\ c,\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\\ \Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}-\dfrac{x}{6}+\dfrac{6x}{6}=0\\ \Leftrightarrow2x-6x-3-x+6x=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)
d, ĐKXĐ:\(x\ne-2,x\ne3\)
\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6}{\left(x+2\right)\left(3-x\right)}+\dfrac{x^2+2x}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{6-2x}{\left(x+2\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{-x^2+x+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\\ \Rightarrow0=0\left(luôn.đúng\right)\)
Y
12 tháng 1 2023
\(a,\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(c,\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(e,\left(x-4\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
\(f,\left(2x-1\right)\left(3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
H
12 tháng 1 2023
`a,(x-1)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
`b,(x -2)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
`c,(x +3)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
`d,(x + 1/2)(4x + 4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
`e,(x -4)(5x -10)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
`f,(2x -1)(3x +6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`g,(2,3x -6,9)(0,1x -2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)
14 tháng 1 2016
a)x=-17
b)x=9/10
c)x=4\(\frac{1}{3}\)
tick đi giải chi tiết cho
14 tháng 1 2016
a)Sử dụng tính chất tỉ lệ thức, có thể biến đổi phương trình như sau
7x+35/3=2x+6/1=>(7x+35)1=3(2x+6)
=>x=-17
b)Sử dụng tính chất tỉ lệ thức, có thể biến đổi phương trình như sau
17x+19/20=27x+10/20=>(17x+19)20=20(27x+10)
c)<=>(x-2)^3+(x-4)^3+(x-7)^3+(-3)(x-2)(x-4)(x-7)=19(3x-13)
=>19(3x-13)=0
rút gọn 57x=247
=>19.3x=19.13
=>3x=13
=>x=13/3
=>x=4\(\frac{1}{3}\)
9 tháng 4 2017
1/ y(y+4)=21 -> y^2 +4y -21=0 -> (y-3)(y+7)=0
VẬY y=3, -7.
2/???
3/(y-4)(y-1)=0 -> y=4, 1
THOI, MAY CAI CO BAN SGK CUNG HOI.DẸP, TỰ LÀM NỐT ĐI, DỄ MÀ.
XONG BẤM ĐÚNG CHO MÌNH
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=0\)
\(\Leftrightarrow x\left(x+10\right)\left(x+4\right)\left(x+6\right)+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)\left(x^2+10x+24\right)+128=0\)
Đặt \(x^2+10x+12=t\)
\(\Rightarrow\left(t-12\right)\left(t+12\right)+128=0\)
\(\Leftrightarrow t^2-144+128=0\)\(\Leftrightarrow t^2-16=0\)
\(\Leftrightarrow\left(t-4\right)\left(t+4\right)=0\)\(\Leftrightarrow\left(x^2+10x+12-4\right)\left(x^2+10x+12+4\right)=0\)
\(\Leftrightarrow\left(x^2+10x+8\right)\left(x^2+10x+16\right)=0\)
\(\Leftrightarrow\left(x^2+10x+8\right)\left(x+2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-8;-2\right\}\)
Ta có : \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)\left(x^2+10x+24\right)+128=0\) (2)
Đặt \(x^2+10x=t\) Khi đó pt (2) có dạng :
\(t\cdot\left(t+24\right)+128=0\)
\(\Leftrightarrow t^2+24t+128=0\)
\(\Leftrightarrow\left(t+12\right)^2-16=0\)
\(\Leftrightarrow\left(t+12-4\right)\left(t+12+4\right)=0\)
\(\Leftrightarrow\left(t+8\right)\left(t+16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t+8=0\\t+16=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}t=-8\\t=-16\end{cases}}\)
+) Với \(t=-8\) thì \(x^2+10x=-8\)
\(\Leftrightarrow\left(x+5\right)^2=17\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=\sqrt{17}\\x+5=-\sqrt{17}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-5+\sqrt{17}\\x=-5-\sqrt{17}\end{cases}}\) ( thỏa mãn )
+) Với \(t=-16\) thì \(x^2+10x=-16\)
\(\Leftrightarrow\left(x+5\right)^2-9=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+14\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+14=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-14\end{cases}}\) ( thỏa mãn )
Vậy : phương trình đã cho có tập nghiệm \(S=\left\{-5\pm\sqrt{17},4,-14\right\}\)