\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.

b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.

Lời giải:

a.

$|4x-1|-|3x-\frac{1}{2}|=0$

$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)

b. Nếu $x\geq 1$ thì:

$|x-1|-2x=\frac{1}{2}$

$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$

$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)

Nếu $x< 1$ thì:

$1-x-2x=\frac{1}{2}$

$\Leftrightarrow x=\frac{1}{6}$ (tm)

\(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\\ \Leftrightarrow x^2+6x+9-x^2+4=4x+17\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\\ \Leftrightarrow x^2+6x+9-x^2+4=4x+17\\ \Leftrightarrow x^2-x^2+6x-4x=17-4-9\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=\dfrac{4}{2}=2\)

x = 2 bạn ạ

P(x)=2x^4-x-2x^3+1

Q(x)=5x^2-x^3+4x-8

tính h(x)=P+Q

 h(x)=P+Q=2x^4-x-2x^3+1+5x^2-x^3+4x-8

\(=2x^4-3x^3+5x^2+3x-7\)

k(x)=Q-P=2x^4-x-2x^3+1-(5x^2-x^3+4x-8)

\(=2x^4-2x^3-x+1-5x^2+x^3-4x+8\)

\(=2x^4-x^3-5x^2-5x+9\)

ng công tỉnh lm nhầm câu b rồi nhưng thôi cx c.

ơn bn

Bn thông cảm.Bài này mn ko bt làm

a)x⁷+x⁵+1=x⁷+x⁶-x⁶+2x⁵-x⁵+x⁴-x⁴+x³-x³+x²-x²+1

=x⁷-x⁶+x⁵-x³+x²+x⁶-x⁵+x⁴-x²+x+x⁵-x⁴+x³-x+1

=x²(x⁵-x⁴+x³-x+1)+x(x⁵-x⁴+x³-x+1)+1(x⁵-x⁴+x³-x+1)

=(x²+x+1)(x⁵-x⁴+x³-x+1)

b)4x⁴-32x²+1=4x⁴+12x³+2x²-12x³-36x²-6x+2x²+6x+1

=2x²(2x²+6x+1)-6x(2x²+6x+1)+1(2x²+6x+1)

=(2x²-6x+1)(2x²+6x+1)

c)x⁶+27=(x²+3)(x²-3x+3)(x²+3x+3)

d)3(x⁴+x²+1)-(x²+x+1)

=3x⁴-3x³+2x²+3x³-3x²+2x+3x²-3x+2

=x²(3x²-3x+2)+x(3x²-3x+2)+1(3x²-3x+2)

=(x²+x+1)(3x²-3x+2)

e)bạn tự làm nhé

Chị cũng đang ko biét ở bài này nè. Hu hu

a, Ta có : \(-x^2+2x-1-3\)

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

Ta thấy : \(\left(x-1\right)^2\ge0\forall x\)

=> \(-\left(x-1\right)^2-3\le-3\forall x\)

Vậy Max = -3 <=> x = 1 .

b, Ta có : \(-x^2-4x-4+4\)

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

Ta thấy : \(\left(x+2\right)^2\ge0\forall x\)

=> \(-\left(x+2\right)^2+4\le4\forall x\)

Vậy Max = 4 <=> x = -2 .

c, Ta có : \(-9x^2+24x-16-2\)

\(=-9\left(x^2-\frac{2.4x}{3}+\frac{16}{9}\right)-2\)

\(=-9\left(x-\frac{4}{3}\right)^2-2\)

Ta thấy : \(\left(x-\frac{4}{3}\right)^2\ge0\forall x\)

=> \(-9\left(x-\frac{4}{3}\right)^2-2\le-2\forall x\)

Vậy Max = -2 <=> x = \(\frac{4}{3}\) .

d, Ta có : \(-x^2+4x-4+3\)

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

Ta thấy : \(\left(x-2\right)^2\ge0\forall x\)

=> \(-\left(x-2\right)^2+3\le3\forall x\)

Vậy Max = 3 <=> x = 2 .

e, Ta có : \(-x^2+2x-1-4y^2-4y-1+7\)

\(=-\left(x-1\right)^2-4\left(y^2+y+\frac{1}{4}\right)+7\)

\(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2+7\)

Ta thấy : \(\left\{{}\begin{matrix}\left(x-1\right)^2\\\left(y+\frac{1}{2}\right)^2\end{matrix}\right.\) \(\ge0\forall xy\)

=> \(\left\{{}\begin{matrix}-\left(x-1\right)^2\\-4\left(y+\frac{1}{2}\right)^2\end{matrix}\right.\) \(\le0\forall xy\)

=> \(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2\le0\forall xy\)

=> \(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2+7\le7\forall xy\)

Vậy Max = 7 <=> \(\left\{{}\begin{matrix}x=1\\y=-\frac{1}{2}\end{matrix}\right.\)

\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy dấu \("="\) không xảy ra