a: ĐKXĐ: \(\begin{cases}x-2\ge0\\ x+1\ge0\\ x^2-x-2\ge0\end{cases}\Rightarrow\begin{cases}x\ge2\\ x\ge-1\\ \left(x-2\right)\left(x+1\right)\ge0\end{cases}\)

=>x>=2

Ta có: \(3x + 14 + 5\sqrt{x-2} = 7\left(\sqrt{x+1} + \sqrt{x^2 - x - 2}\right)\) (1)

Đặt \(a=\sqrt{x-2};b=\sqrt{x+1}\) (ĐIều kiện: a>=0; b>0)

=>\(ab=\sqrt{\left(x-2\right)\left(x+1\right)}=\sqrt{x^2-x-2}\)

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

\(b^2-1=x+1-1=x\)

=>\(3x+14=3\left(b^2-1\right)+14=3b^2+11\)

(1) sẽ tương đương: \(3b^2+11+5a=7b+7ab\)

=>\(3b^2-7b+11+a(5-7b)=0\)

=>\((b - 2)(40b^3 + 52b^2 - 133b + 98) = 0\)

=>b-2=0

=>b=2

=>x+1=4

=>x=3

b: ĐKXĐ: 7/3<=x<=7

\(7\sqrt{3x-7}+(4x-7)\sqrt{7-x}=32\left(2\right)\)

Đặt \(a=\sqrt{3x-7};b=\sqrt{7-x}\)

=>\(a^2+b^2=3x-7+7-x=2x\)

\(3a^2+b^2=3\left(3x-7\right)+7-x=9x-21+7-x=8x-14\)

\(3x-7=a^2\)

=>\(3x=a^2+7\)

=>\(x=\frac{a^2+7}{3}\)

=>\(4x-7=4\cdot\frac{a^2+7}{3}-7=\frac{4\left(a^2+7\right)-21}{3}=\frac{4a^2+7}{3}\)

(2) sẽ trở thành: \(7a+\frac{4a^2+7}{3}b=32\iff21a+(4a^2+7)b=96\)

mà \(a^2 + 3b^2 = 14\)

nên \(a=\sqrt5;b=\sqrt3\)

=>3x-7=5

=>3x=12

=>x=4(nhận)

a) Thay \(x=2\) vào phương trình, ta được:

 \(15\left(m+6\right)+12=80\) \(\Rightarrow m=-\dfrac{22}{15}\)

Vậy \(m=-\dfrac{22}{15}\)

b) Thay \(x=1\) vào phương trình, ta được:

  \(15\left(2+m\right)-32=43\) \(\Rightarrow m=3\)

Vậy \(m=3\)

a,\(11-2x=x-1\Leftrightarrow-2x-x=-1-11\Leftrightarrow-3x=-12\Leftrightarrow x=-4\)

b,\(\text{5(3x+2)=4x+1}\Leftrightarrow15x+10=4x+1\Leftrightarrow15x-4x=1-10\Leftrightarrow11x=-9\Leftrightarrow x=\dfrac{-9}{11}\)

c,\(x^2-4-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x+2\right)\left(x-2\right)-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x-2\right)[\left(x+2\right)-\left(x-5\right)]\Leftrightarrow\left(x-2\right)\left[x+2-x+5\right]\Leftrightarrow\left(x-2\right)7\Leftrightarrow7x-14\)

Câu 1: ĐKXĐ: ...

\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)

\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)

\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)

\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)

\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow...\)

Câu 2:

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)

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

\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)

\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)

\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)

\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)

\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)

\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)

\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)

Tìm được mỗi nghiệm thôi à :v

a: \(\Leftrightarrow x+2-3xm-m=5\)

\(\Leftrightarrow x\left(1-3m\right)=5+m-2=m+3\)

Để đây là pt bậc nhất thì -3m+1<>0

hay m<>1/3

b: Khi m=-1 thì pt sẽ là \(x\left(1+3\right)=-1+3=2\)

=>x=1/2

Giải pt :

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

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

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

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

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

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

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

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

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

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

Tập nghiệm của pt là : \(S=\left\{-2;\dfrac{1}{4}\right\}\)

cam on

Ta có:

$\dfrac{x^4}{2x^2+1}+\dfrac{2x^2+1}{x^4}=2$

Điều kiện: $x\ne0$.

Đặt $t=\dfrac{x^4}{2x^2+1}>0$.

Khi đó:

$t+\dfrac{1}{t}=2$

$t^2-2t+1=0$

$(t-1)^2=0$

$t=1$

$\dfrac{x^4}{2x^2+1}=1$

$x^4=2x^2+1$

$x^4-2x^2-1=0$

Đặt $y=x^2\ge0$:

$y^2-2y-1=0$

$y=1+\sqrt2$ (do $y\ge0$)

$x^2=1+\sqrt2$

$x=\pm\sqrt{1+\sqrt2}$

Vậy $x=\pm\sqrt{1+\sqrt2}$.

Ta có:

$\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}$

Điều kiện: $x\ne\pm1$.

Quy đồng:

$\dfrac{x^2(x+1)^2+x^2(x-1)^2}{(x^2-1)^2}=\dfrac{10}{9}$

$\dfrac{x^2[(x+1)^2+(x-1)^2]}{(x^2-1)^2}=\dfrac{10}{9}$

$\dfrac{2x^2(x^2+1)}{(x^2-1)^2}=\dfrac{10}{9}$

$18x^2(x^2+1)=10(x^2-1)^2$

$9x^2(x^2+1)=5(x^2-1)^2$

$9x^4+9x^2=5x^4-10x^2+5$

$4x^4+19x^2-5=0$

Đặt $t=x^2\ge0$:

$4t^2+19t-5=0$

$(4t-1)(t+5)=0$

$t=\dfrac14$ (do $t\ge0$)

$x^2=\dfrac14$

$x=\pm\dfrac12$

Vậy $x=\pm\dfrac12$.

Ta có:

$x^3+3x^2-10x-24=0$

Nhóm các hạng tử:

$x^2(x+3)-10(x+3)=0$

$(x+3)(x^2-10)=0$

$x+3=0$ hoặc $x^2-10=0$

$x=-3$ hoặc $x=\pm\sqrt{10}$

Vậy $x\in{-3,-\sqrt{10},\sqrt{10}}$.

a)               \(3x^2+x-4=0\)

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=-\frac{4}{3}\end{cases}}\)

Vậy..

b)    \(2x^2-x-28=0\)

\(\Leftrightarrow\)\(\left(x-4\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=4\\x=-3.5\end{cases}}\)

Vậy...

c)     \(6x^2-x-7=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{6}\end{cases}}\)

Vậy....

d)  \(3x^2-5=0\)

\(\Leftrightarrow\)\(3x^2=5\)

\(\Leftrightarrow\)\(x^2=\frac{5}{3}\)

\(\Leftrightarrow\)\(x=\pm\sqrt{\frac{5}{3}}\)

Vậy...

=>\(\dfrac{-1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}=2\)

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

=>\(\dfrac{x-1-x+4}{x^2-5x+4}=2\)

=>2x^2-10x+8=3

=>2x^2-10x+5=0

=>\(x=\dfrac{5\pm\sqrt{15}}{2}\)