1.

A =\(2x^2-8x+10=\left(x^2-2x+1\right)+\left(x^2-6x+9\right)\)

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

Có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(3-x\right)^2\ge0\end{matrix}\right.\forall x\)

<=> \(\left|x-1\right|+\left|x-3\right|\)

Áp dụng bđt |a| + |b| \(\ge\) |a + b| có:

\(\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)

đẳng thức xảy ra khi \(1\le x\le3\)

Vậy ................

\(2ab^2-a^2b-b^3=b\left(2ab-a^2-b^2\right)\)

\(=b\left[-\left(a^2-2ab+b^2\right)\right]=b\left[-\left(a-b\right)^2\right]=-b\left(a-b\right)^2\)

a) \(\left(2x-1\right)^4=81\)

\(\Rightarrow\left(2x-1\right)^4=\left(\pm3\right)^4\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy............

b) \(\left(x-1\right)^5=-32\)

\(\Rightarrow\left(x-1\right)^5=-2^5\)

\(\Rightarrow x-1=-2\Rightarrow x=-1\)

Vậy.......

c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Rightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)

\(\Rightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)

Vậy x = 1/2; x = 1; x = 0

1, \(x^4-19x^2-10x+8=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3-4x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+1\right)\left(x^2-5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\\x^2-5x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x_1=-4\\x_2=-1\end{matrix}\right.\)

hoặc \(x^2-5x+2=0\)

\(\Rightarrow\Delta=17\left(CT:b^2-4ac\right)\)

\(\Rightarrow\left[{}\begin{matrix}x_3=\dfrac{5+\sqrt{17}}{2}\\x_4=\dfrac{5-\sqrt{17}}{2}\end{matrix}\right.\)

Vậy pt có 4 n^o là...........

\(\dfrac{\left(x-2\right)}{\left(x-1\right)}=\dfrac{\left(x+4\right)}{\left(x+7\right)}\)

\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)

\(\Rightarrow x^2+7x-2x-14=x^2-x+4x-4\)

\(\Rightarrow x^2+5x-14=x^2+3x-4\)

\(\Rightarrow x^2+5x-x^2-3x=-4+14\)

\(\Rightarrow2x=10\Rightarrow x=\dfrac{10}{2}=5\)

Vậy......

thằng này nứng ak /-/ mi thik ns xấu ai đò t thế ak /-/ hay mi gato vại ==

1. a, | 2x - 3 | + x = 5

<=> | 2x - 3| = 5 - x

<=> \(\left[{}\begin{matrix}2x-3=5-x\\2x-3=-5+x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=8\Rightarrow x=\dfrac{8}{3}\\x=-2\end{matrix}\right.\)

b, 3x - 2 +2| x + 3| = 0

Với x \(\ge1\) có:

3x - 2 + 2x + 6 = 0

<=> 5x = -4

<=> \(x=-\dfrac{4}{5}\)

Với x < 1 có:

-3x - 2 - 2x + 6 = 0

<=> -5x = -4

<=> x = \(\dfrac{4}{5}\) thử lại k thỏa mãn

Vậy có 1 gt x tm đề là x = -4/5

c, Tương tự b

Bài 2: gần tương tự bài 1

Bài 3:

a, Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:

\(\left|x\right|+\left|8-x\right|\ge\left|x+8-x\right|=8\)

đẳng thắc xảy ra khi \(0\le x\le8\)

Vậy A_min = 8 khi.....

b, Áp dụng bđt như ý a ta có:

\(\left|x-2\right|+\left|5-x\right|\ge\left|x-2+5-x\right|=3\)

đẳng thức xảy ra khi \(2\le x\le5\)

Vậy...............