a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

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

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

a.Ta có:|2x-1|=2x-1\(\Leftrightarrow\)2x-1\(\ge\)0\(\Leftrightarrow\)x\(\ge\)\(\dfrac{1}{2}\)

|2x-1|=1-2x\(\Leftrightarrow\)2x-1<0\(\Leftrightarrow\)x<\(\dfrac{1}{2}\)

ĐK:\(x\ge\dfrac{1}{2}\)

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

\(\Leftrightarrow2x-1-2x+1=0\)

\(\Leftrightarrow0x=0\)

\(\Rightarrow\)Tập no của PT là S={\(\forall x\)|x\(\ge\dfrac{1}{2}\)}

b.|0,5-3x|=3x-0,5\(\Leftrightarrow\)x<2,5

=0,5-3x\(\Leftrightarrow x\ge2,5\)

ĐK:x<2,5

Gỉai

0,5-3x=3x-0,5

\(\Leftrightarrow\)0,5-3x-3x+0,5=0

\(\Leftrightarrow\)1-6x=0

\(\Leftrightarrow x=\dfrac{1}{6}\)(TMĐKXĐ)

\(\Rightarrow\)tập no của PT là S={\(\dfrac{1}{6}\)}

c.|5x+1-10x|=0,5\(\Leftrightarrow\)|1-5x|=0,5\(\Leftrightarrow x< \dfrac{1}{5}\)

\(\Leftrightarrow\)|1-5x|=-0,5\(\Leftrightarrow\)x\(\ge\dfrac{1}{5}\)

ĐK:\(x< \dfrac{1}{5}\)

Gỉai

1-5x=0,5

\(\Leftrightarrow5x=0,5\)

\(\Leftrightarrow x=0,1\)(loại)

\(\Rightarrow pt\) trên vô nghiệm

d.|x+2|-|x-7|=0

ĐK:x\(\ne\pm2\);x\(\ne\pm7\)

Gỉai

\(\left\{{}\begin{matrix}x+2-x+7=0\\x-2-x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-9=0\\-2x-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-9=0\left(KTMĐKXĐ\right)\\x=-4,5\left(TMĐKXĐ\right)\end{matrix}\right.\)

\(\Rightarrow\)tập no của phương trình là S={-4,5}

a, \(\left|x+2\right|-\left|x+7\right|=0\Rightarrow\left|x+2\right|=\left|x+7\right|\Rightarrow\orbr{\begin{cases}x+2=x+7\\x+2=-x-7\end{cases}\Rightarrow\orbr{\begin{cases}0=5\left(loại\right)\\2x=-9\end{cases}\Rightarrow}x=\frac{-9}{2}}\)

b, - Nếu \(2x-1\ge0\Rightarrow x\ge\frac{1}{2}\), ta có: 2x - 1 = 2x - 1 => 2x = 2x (thỏa mãn với mọi x)

- Nếu 2x - 1 < 0 => \(x< \frac{1}{2}\), ta có: 2x - 1 = 1 - 2x => 4x = 2 => x = \(\frac{1}{2}\) (không thỏa mãn điều kiện)

Vậy \(x\ge\frac{1}{2}\)

c,d tương tự b

e, tương tự a

Bài 1:\(\left|2x-1\right|=2x-1\) khi \(x>0\)

b)\(\left|0,5-3x\right|=3x-0.5\) khi x= 4

c)\(\left|5x+1\right|-10x=0,5\) khi x= 0,1

Bài 2:Min A=0

Min B=-2

Bài 1:

a, \(\left|2x-1\right|=2x-1\)

+) Xét \(x\ge\dfrac{1}{2}\) ta có:
\(2x-1=2x-1\)

\(\Rightarrow x\) tùy ý với \(x\ge\dfrac{1}{2}\)

+) Xét \(x< \dfrac{1}{2}\) ta có:

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

\(\Rightarrow4x=2\)

\(\Rightarrow x=\dfrac{1}{2}\) ( không t/m )

Vậy...

b, \(\left|0,5-3x\right|=3x-0,5\)

+) Xét \(x\ge\dfrac{1}{6}\) ta có:

\(0,5-3x=3x-0,5\)

\(\Rightarrow6x=1\)

\(\Rightarrow x=\dfrac{1}{6}\) ( t/m )
+) Xét \(x< \dfrac{1}{6}\) ta có:

\(3x-0,5=3x-0,5\)

\(\Rightarrow x\) tùy ý với \(x< \dfrac{1}{6}\)

Vậy \(x\le\dfrac{1}{6}\)

c, \(\left|5x+1\right|-10x=0,5\)

+) Xét \(x\ge\dfrac{-1}{5}\) ta có:

\(5x+1-10x=0,5\)

\(\Rightarrow-5x=-0,5\)

\(\Rightarrow x=\dfrac{1}{10}\) ( t/m )

+) Xét \(x< \dfrac{-1}{5}\) ta có:

\(-5x-1-10x=0,5\)

\(\Rightarrow-15x=1,5\)

\(\Rightarrow x=\dfrac{-1}{10}\) ( không t/m )

Vậy \(x=\dfrac{1}{10}\)

Bài 2:

a, Ta có: \(-\left|x-3,5\right|\le0\)

\(\Rightarrow A=0,5-\left|x-3,5\right|\le3,5\)

Dấu " = " xảy ra khi \(-\left|x-3,5\right|=0\Rightarrow x=3,5\)

Vậy \(MIN_A=0,5\) khi x = 3,5

b, Ta có: \(-\left|1,4-x\right|\le0\)

\(\Rightarrow B=-\left|1,4-x\right|-2\le-2\)

Dấu " = " xảy ra khi \(-\left|1,4-x\right|=0\Rightarrow x=1,4\)

Vậy \(MIN_B=-2\) khi \(x=1,4\)

Bài 2: tìm gtln ms đúng nhé!

a/Có: \(-\left|x-3,5\right|\le0\forall x\Rightarrow0,5-\left|x-3,5\right|\le0,5\)

Dấu ''='' xảy ra \(\Leftrightarrow x=3,5\)

Vậy \(A_{MAX}=0,5\Leftrightarrow x=3,5\)

b, có: \(-\left|1,4-x\right|\le0\Rightarrow-\left|1,4-x\right|-2\le-2\)

Dấu ''='' xảy ra \(\Leftrightarrow x=1,4\)

Vậy \(B_{MAX}=-2\Leftrightarrow x=1,4\)

a) Ta có: \(\left(2x+1\right)^2-\left(3x-4\right)^2=0\)

\(\Leftrightarrow\left(2x+1-3x+4\right)\left(2x+1+3x-4\right)=0\)

\(\Leftrightarrow\left(5-x\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{5}\end{matrix}\right.\)

b) Ta có: \(5x^3-3x^2+10x-6=0\)

\(\Leftrightarrow x^2\left(5x-3\right)+2\left(5x-3\right)=0\)

\(\Leftrightarrow5x-3=0\)

hay \(x=\dfrac{3}{5}\)

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

a: =>3x^2-3x-2x+2=0

=>(x-1)(3x-2)=0

=>x=2/3 hoặc x=1

b: =>2x^2=11

=>x^2=11/2

=>\(x=\pm\dfrac{\sqrt{22}}{2}\)

c: Δ=5^2-4*1*7=25-28=-3<0

=>PTVN

f: =>6x^4-6x^2-x^2+1=0

=>(x^2-1)(6x^2-1)=0

=>x^2=1 hoặc x^2=1/6

=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)

d: =>(5-2x)(5+2x)=0

=>x=5/2 hoặc x=-5/2

e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2

=>3x^2+5x-8=0 và x>=-1/2

=>3x^2+8x-3x-8=0 và x>=-1/2

=>(3x+8)(x-1)=0 và x>=-1/2

=>x=1