`|5x| = - 3x + 2`

Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :

`5x =-3x+2`

`<=> 5x +3x=2`

`<=> 8x=2`

`<=> x= 2/8=1/4` ( thỏa mãn )

Nếu `5x<0<=>x<0` thì phương trình trên trở thành :

`-5x = -3x+2`

`<=>-5x+3x=2`

`<=> 2x=2`

`<=>x=1` ( không thỏa mãn ) 

Vậy pt đã cho có nghiệm `x=1/4`

__

`6x-2<5x+3`

`<=> 6x-5x<3+2`

`<=>x<5`

Vậy bpt đã cho có tập nghiệm `x<5`

Bài 1: Giải các phương trình sau:

a) 3(2,2-0,3x)=2,6 + (0,1x-4)

<=> 6.6 - 0.9x = 2,6 + 0,1x - 4

<=> - 0.9x - 0,1x = -6.6 -1,4

<=> -x = -8

<=> x = 8

Vậy x = 8

b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)

<=> 3,6 - x - 0,5 = x - 5,5 + x

<=> - x - 3,1 = -5,5

<=> - x = -2.4

<=> x = 2.4

Vậy  x = 2.4

1.Thay m=-1 vào pt ta được:

\(x^4-2x^2-3=0\)\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vn\right)\\x^2=3\end{matrix}\right.\)\(\Rightarrow x=\pm\sqrt{3}\)

Vậy...

2.Đặt \(t=x^2\left(t\ge0\right)\)

Với mỗi t>0 thì sẽ luôn có hai x phân biệt

Pttt: \(t^2-2t+m-2=0\) (2)

Để pt (1) có 4 nghiệm pb \(\Leftrightarrow\) PT (2) có hai nghiệm pb dương

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\S=2>0\left(lđ\right)\\P=m-2>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}4-4\left(m-2\right)>0\\m>2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m< 3\\m>2\end{matrix}\right.\)\(\Rightarrow2< m< 3\)

Vậy...

1. Bạn tự giải

2. Đặt \(x^2=t\ge0\) pt trở thành:

\(t^2-2t+m-2=0\) (2)

Pt đã cho có 4 nghiệm pb khi và chỉ khi (2) có 2 nghiệm dương pb

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta'=1-\left(m-2\right)>0\\t_1+t_2=2>0\\t_1t_2=m-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m< 3\\m>2\end{matrix}\right.\)

\(\Rightarrow2< m< 3\)

`4)(2x-5)/5-(x+3)/3=(2-3x)/2-x-2`

`<=>6(2x-5)-10(x+3)=15(2-3x)-30x-60`

`<=>12x-30-10x-30=30-45x-30x-60`

`<=>2x-60=-30-75x`

`<=>77x=30`

`<=>x=30/77`

Vậy `S={30/77}`

`12)(x^2-3x)^2-2(x^2-3)=8`

`<=>x^4+9x^2-6x^3-2x^2+6-8=0`

`<=>x^4-6x^3+7x^2-2=0`

`<=>x^4-x^3-5x^3+5x^2+2x^2-2x+2x-2=0`

`<=>x^3(x-1)-5x^2(x-1)+2x(x-1)+2(x-1)=0`

`<=>(x-1)(x^3-5x^2+2x+2)=0`

`<=>(x-1)(x^3-x^2-4x^2+4x-2x+2)=0`

`<=>(x-1)[x^2(x-1)-4x(x-1)-2(x-1)]=0`

`<=>(x-1)^2(x^2-4x-2)=0`

`<=>(x-1)^2[(x-2)^2-6]=0`

`<=>(x-1)(x-2-\sqrt{6})(x-2+\sqrt{6})=0`

`<=>` $\left[ \begin{array}{l}x=1\\x=2-\sqrt{6}\\x=2+\sqrt{6}\end{array} \right.$

a) Ta có: \(3x-\dfrac{1}{2}+5\left(x-2\right)=\dfrac{2}{3}\left(x+1\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}+5x-10=\dfrac{2}{3}x+\dfrac{2}{3}\)

\(\Leftrightarrow8x-\dfrac{21}{2}-\dfrac{2}{3}x-\dfrac{2}{3}=0\)

\(\Leftrightarrow\dfrac{22}{3}x-\dfrac{67}{6}=0\)\(\Leftrightarrow x\cdot\dfrac{22}{3}=\dfrac{67}{6}\)

\(\Leftrightarrow x=\dfrac{67}{6}:\dfrac{22}{3}=\dfrac{67}{6}\cdot\dfrac{3}{22}=\dfrac{67}{44}\)

Vậy: \(S=\left\{\dfrac{67}{44}\right\}\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

c) Ta có: \(2\left(x-3\right)+1=2\left(x+1\right)-9\)

\(\Leftrightarrow2x-6+1=2x+2-9\)

\(\Leftrightarrow-5=-7\)(Vô lý)

Vậy: \(S=\varnothing\)

d) Ta có: \(\dfrac{7x}{8}-5\left(x-9\right)=\dfrac{20x+1.5}{6}\)

\(\Leftrightarrow\dfrac{21x}{24}-\dfrac{120\left(x-9\right)}{24}=\dfrac{4\left(20x+1.5\right)}{24}\)

\(\Leftrightarrow21x-120x+1080=80x+6\)

\(\Leftrightarrow-99x+1080-80x-6=0\)

\(\Leftrightarrow-179x+1074=0\)

\(\Leftrightarrow-179x=-1074\)

hay x=6

Vậy: S={6}

f) Ta có: \(3\left(x+4\right)-5\left(x-2\right)=4\left(3x-1\right)+82\)

\(\Leftrightarrow3x+12-5x+10=12x-4+82\)

\(\Leftrightarrow-2x+22-12x-78=0\)

\(\Leftrightarrow-14x-56=0\)

\(\Leftrightarrow-14x=56\)

hay x=-4

Vậy: S={-4}

g) Ta có: \(\dfrac{x+17}{5}-\dfrac{3x-7}{4}=-2\)

\(\Leftrightarrow\dfrac{4\left(x+17\right)}{20}-\dfrac{5\left(3x-7\right)}{20}=\dfrac{-40}{20}\)

\(\Leftrightarrow4x+68-15x+35+40=0\)

\(\Leftrightarrow-11x+143=0\)

\(\Leftrightarrow-11x=-143\)

hay x=13

Vậy: S={13}

a) \(3x-\dfrac{1}{2}+5\left(x-2\right)=\dfrac{2}{3}\left(x+1\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}+5x-10=\dfrac{2}{3}x+\dfrac{2}{3}\)

\(\Leftrightarrow3x+5x-\dfrac{2}{3}x=\dfrac{1}{2}+10+\dfrac{2}{3}\)

\(\Leftrightarrow\left(3+5-\dfrac{2}{3}\right)x=\dfrac{67}{6}\)

\(\Leftrightarrow\dfrac{22}{3}x=\dfrac{67}{6}\)

\(\Leftrightarrow x=\dfrac{67}{44}\)

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

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

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

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

 \(\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

c) \(2\left(x-3\right)+1=2\left(x+1\right)-9\)

\(\Leftrightarrow2x-6+1=2x+2-9\)

\(\Leftrightarrow2x-5=2x-7\)

 => x vô nghiệm

d và e giống nhau) \(\dfrac{7x}{8}-5\left(x-9\right)=\dfrac{20x+1,5}{6}\)

\(\Leftrightarrow\dfrac{7x}{8}-5x+45=\dfrac{20x+1,5}{6}\)

\(\Leftrightarrow\dfrac{7x}{8}-\dfrac{40x}{8}+\dfrac{360}{8}=\dfrac{20x+1,5}{6}\)

\(\Leftrightarrow\dfrac{-33x+360}{8}=\dfrac{20x+1,5}{6}\)

\(\Leftrightarrow\dfrac{-99x+1080}{24}=\dfrac{80x+6}{24}\)

\(\Leftrightarrow-99x+1080=80x+6\)

\(\Leftrightarrow-179x=-1074\)

\(\Leftrightarrow x=6\)

f) \(3\left(x+4\right)-5\left(x-2\right)=4\left(3x-1\right)+82\)

\(\Leftrightarrow3x+12-5x+10=12x-4+82\)

\(\Leftrightarrow-2x+22=12x+78\)

\(\Leftrightarrow-14x=56\)

\(\Leftrightarrow x=-4\)

g) \(\dfrac{x+17}{5}-\dfrac{3x-7}{4}=-2\)

\(\Leftrightarrow\dfrac{x+17}{5}-\dfrac{3x-7}{4}+2=0\)

\(\Leftrightarrow\dfrac{4x+68-15x+35+40}{20}=\dfrac{0}{20}\)

\(\Leftrightarrow4x+68-15x+35+40=0\)

\(\Leftrightarrow-11x+143=0\)

\(\Leftrightarrow-11x=143\)

\(\Leftrightarrow x=-13\)

`x(x-1)(x+1)(x+2)=24`

`<=>[x(x+1)][(x-1)(x+2)]=24`

`<=>(x^2+x)(x^2+x-2)=24`

`<=>(x^2+x-1)^2=25`

`+)x^2+x-1=5`

`<=>x^2+x-6=0`

`<=>x^2-2x+3x-6=0`

`<=>x(x-2)+3(x-2)=0`

`<=>(x-2)(x+3)=0`

`<=>` $\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.$

`+)x^2+x-1=-5`

`<=>x^2+x+4=0`

`<=>(x+1/2)^2+15/4=0` vô lý

Vậy `S={2,3}`

`11)4x^2+4-8x=9(x-2)^2`

`<=>4(x^2-2x+1)=9(x-2)^2`

`<=>(2x-2)^2=(3x-6)^2`

`<=>(3x-6-2x+2)(3x-6+2x-2)=0`

`<=>(x-4)(5x-8)=0`

`<=>` $\left[ \begin{array}{l}x=4\\x=\dfrac{8}{5}\end{array} \right.$

Vậy `S={4,5/8}`

\(a,PT\Leftrightarrow\left(1-2m\right)x=m+4\)

Bậc nhất \(\Leftrightarrow1-2m\ne0\Leftrightarrow m\ne\dfrac{1}{2}\)

\(b,x=2\Leftrightarrow2-4m-m-4=0\Leftrightarrow m=-\dfrac{2}{5}\\ c,m=5\Leftrightarrow-9x-9=0\Leftrightarrow x=-1\)

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