Câu 10: A

Câu 11: A

Câu 12: C

Câu 13: A

Câu 15: B

Câu 16: C

Câu 17: B

Câu 18: D

a, x\(^2\) - x = x - 1
\(\Leftrightarrow\) x\(^2\) - 2x + 1 = 0
\(\Leftrightarrow\) (x - 1)\(^2\) = 0
\(\Leftrightarrow\) x - 1 = 0
\(\Leftrightarrow\) x = 1

a) \(x^2-x=x-1\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Rightarrow x=1\)

b) \(\left(x^2-36\right)-\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy..

c) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

\(\Leftrightarrow-4x+2=0\)

\(\Rightarrow x=\dfrac{1}{2}\)

d) \(x^2\left(x^2-4\right)-\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm1\end{matrix}\right.\)

Vậy..

a) x^2 - x = x - 1

\(x^2-x-\left(x-1\right)=0\)

\(x^2-2.x.1+1^2=0\)

\(\left(x-1\right)^2=0=>x-1=0=>x=1\)
b) (x^2 - 36) - (x+6) = 0

=> \(\left(x^2-6^2\right)-\left(x+6\right)=0\)

=> (x+6)(x-6) -(x+6) =0

=> (x-6)(x+6-1) =0

=> (x-6)(x+5)=0

=> x=6 hoặc x= (-5)
c) (2x-1)^2 - (4x^2 - 1)= 0

=> \(\left(2x-1\right)^2-\left(\left(2x\right)^2-1^2\right)=0\)

=>\(\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)

=> (2x-1)(2x-1-2x-1)=0

=> (2x-1)(-2)=0

=> 2x-1=0 => 2x=1 => x= \(\dfrac{1}{2}\)
d) x^2(x^2 - 4) - (x^2 - 4 ) = 0

\(\left(x^2-2^2\right)\left(x^2-1\right)=0\)

(x-2)(x+2)(x-1)(x+1)=0

=> x=2;-2;1 hoặc (-1)

Đề nghe cứ sao sao ý (mk góp ý thui đừng ném gạch đá nha)

\(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)+8\)

\(A=\left(x^2+6x\right)\left(x^2+6x+8\right)+8\)

Đặt \(t=x^2+6x\)

\(A=t\left(t+8\right)+8\)

\(A=t^2+8x+16-8\)

\(A=\left(t+4\right)^2-8\ge-8\left(\forall t\right)\)

\("="\Leftrightarrow t=-4\Leftrightarrow x^2+6x+4=0\)\(\Leftrightarrow\orbr{\begin{cases}x=-3-\sqrt{5}\\x=-3+\sqrt{5}\end{cases}}\)

a) Ta có: \(x^2\ge0\forall x\in Q\)

\(y^2\ge0\forall x\in Q\)

\(\Rightarrow x^2+y^2+2014\ge2014\forall x\in Q\)

Dấu giá trị nhỏ nhất của biểu thức là 2014, xảy ra khi \(\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

b, Ta có: \(\left(x+30\right)^2\ge0\forall x\in Q\)

\(\left(y-4\right)^2\ge0\forall x\in Q\)

\(\Rightarrow\left(x+30\right)^2+\left(y-4\right)^2+17\ge17\forall x\in Q\)

Dấu giá trị nhỏ nhất của biểu thức là 17, xảy ra khi \(\left\{{}\begin{matrix}\left(x+30\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-30\\y=4\end{matrix}\right.\)

c, Ta có: \(\left(y-9\right)^2\ge0\forall x\in Q\)

\(\left|x-3\right|\ge0\forall x\in Q\)

\(\Rightarrow\left(y-9\right)^2+\left|x-3\right|^2-1\ge-1\forall x\in Q\)

Dấu giá trị nhỏ nhất của biểu thức là -1 xảy ra khi \(\left\{{}\begin{matrix}\left(y-9\right)^2=0\\\left|x-3\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x=3\end{matrix}\right.\)

ghi đề kiểu này khó nhìn quá

a) \(\left(2-x\right)x^2\le0\)

Ta có: \(\left(2-x\right)x^2=0\Leftrightarrow\left[{}\begin{matrix}x^2=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vì \(x^2\ge0\Rightarrow\left(2-x\right)x^2\Leftrightarrow2-x< 0\Leftrightarrow2< x\)

Vậy ......

b, \(\left(x-7\right)\left(x+3\right)< 0\Leftrightarrow\left[{}\begin{matrix}x-7< 0\\x+3< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 7\\x< -3\end{matrix}\right.\)

Vây........

c, \(\left(x+4\right)\left(x+3\right)\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+4< 0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-4\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -4\\x< -3\end{matrix}\right.\end{matrix}\right.\)

Vậy..............

a/ \(x^2-2x=-1\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\Rightarrow x=1\)

Vậy..............

b/ \(x^2+2x+1=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\Rightarrow x=-1\)

Vậy.......

c/ \(4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)=3x^2\)

\(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)

\(\Leftrightarrow-4x=0\Rightarrow x=0\)

Vậy...................

d/ \(x\left(x-2017\right)-x^2\left(2017-x\right)=0\)

\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)

\(\Leftrightarrow x^3-2016x^2-2017x=0\)

\(\Leftrightarrow x^3+x^2-2017x^2-2017x=0\)

\(\Leftrightarrow x\left(x^2+x\right)-2017\left(x^2+x\right)=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x-2017\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\x-2017=0\Rightarrow x=2017\end{matrix}\right.\)

Vậy pt có 3 nghiệm là.....(tự ghi ra)

\(a,x^2-2x=-1\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

\(b,x^2+2x+1=0\)

\(\Leftrightarrow\left(x+2\right)^2=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

\(c,4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)-3x^2=0\) \(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)

\(\Leftrightarrow-4x=0\Rightarrow x=0\)

\(d,x\left(x-2017\right)-x^2\left(2017-x\right)=0\)

\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)

\(\Leftrightarrow x^3+x^2-2017x-2017=0\)

\(\Leftrightarrow x^2\left(x+1\right)-2017\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2017\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-2107=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=2017\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\\left[{}\begin{matrix}x=\sqrt{2017}\\x=-\sqrt{2017}\end{matrix}\right.\end{matrix}\right.\)

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đù khó thế

tl j z mấy chế , k câu dc đâu :))