(x-1)^x+2=(x-1)^2
giup minh voi
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Tinh bang hai cach:
a. (1/3 + 1/5) x 1/2 b. (1/3 - 1/5) x 1/2
giup minh voi, tra loi minh tick cho nha
a.\(=\dfrac{8}{15}\times\dfrac{1}{2}=\dfrac{4}{15}\)
b.\(=\dfrac{1}{3}\times\dfrac{1}{2}-\dfrac{1}{5}\times\dfrac{1}{2}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\)
\(\left(-x^3.z.y\right).\left(\dfrac{2}{3}.y.x^2\right)^2\)
\(=-x^3.z.y.\dfrac{4}{9}.y^2.x^4\)
\(=-\dfrac{4}{9}x^7.y^3.z\)
`(-x^3zy)(2/3yx^2)^2`
`=-4/9x^3zy.y^2x^4`
`=-4/9x^{3+4}.y^{1+2}z`
`=-4/9x^7y^3z`
\(a,=4x^2+4x+1\\ b,=9-12y+4y^2\\ c,=\dfrac{x^2}{4}-xy+y^2\\ d,=\dfrac{25}{4}-5x+x^2\\ e,=4x^2+32xy+64y^2\\ f,=9x^2-30xy+25y^2\)
1/5+8/15:y=1/2
8/15:y=1/2-1/5
8/15:y=3/10
y=8/15:3/10
y=16/9
Vậy y=16/9
\(\left(-x^2y\right)^3\cdot\dfrac{1}{2}\cdot x^2y^3\cdot\left(-2xy^2z\right)^2\\ =-x^6y^3\cdot\dfrac{1}{2}x^2y^3\cdot4x^2y^4z^2\\ =\left(-1\cdot\dfrac{1}{2}\cdot4\right)\cdot\left(x^6\cdot x^2\cdot x^2\right)\cdot\left(y^3\cdot y^3\cdot y^4\right)\cdot z^2\\ =-2x^{10}y^{10}z^2\)
`đk:x ne 0,-2`
`a)D=(x/(x+2)+(8x+8)/(x^2+2x)-(x+2)/x):((x^2-x-3)/(x^2+2x)+1/x)`
`=((x^2+8x+8-x^2-4x-4)/(x(x+2))):((x^2-x-3+x+2)/(x(x+2)))`
`=(4x+4)/(x(x+2)):(x^2-1)/(x(x+2))`
`=(4x+4)/(x^2-1)(x ne +-1)`
`=4/(x-1)`
`b)x(x-2)-(x-2)=0`
`<=>(x-2)(x-1)=0`
Vì `x ne 1=>x-1 ne 0`
`=>x-2=0<=>x=2`
`=>D=4/(2-1)=4`
`c)D<0`
Mà `4>0`
`=>x-1<0`
`=>x<1`
Kết hợp đkxđ:
`=>x<1,x ne 0,x ne -2`
`d)D=2`
`<=>4/(x-1)=2`
`<=>2/(x-1)=1`
`<=>x-1=2`
`<=>x=3(tm)`
Câu 2 bạn ghi thiếu đề
Câu 1:
\(\Leftrightarrow\left(m^2-3m\right)x+2x< 2-m\)
\(\Leftrightarrow\left(m^2-3m+2\right)x< 2-m\)
BPT đã cho vô nghiệm khi và chỉ khi:
\(\left\{{}\begin{matrix}m^2-3m+2=0\\2-m\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=1\\m=2\end{matrix}\right.\\m\ge2\end{matrix}\right.\) \(\Rightarrow m=2\)
\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)
\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)
Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm
\(\Leftrightarrow6x+2-x=0\)
\(\Leftrightarrow5x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)
Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)
\(T=\dfrac{2x^2-y^2}{2x^2+y^2}=\dfrac{2\left(2k\right)^2-\left(3k\right)^2}{2\left(2k\right)^2+\left(3k\right)^2}=\dfrac{8k^2-9k^2}{8k^2+9k^2}=\dfrac{-k^2}{17k^2}=\dfrac{-1}{17}\)
\(x=\left\{0;1\right\}\)
(x - 1)2 + 2 = (x - 1)2
Đề như vậy hả