a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)

Vậy x = 8 hoặc x = -7

a: Ta có: \(x^4-x^2-56=0\)

\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)

\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)

\(\Leftrightarrow x^2-8=0\)

hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)

\(a,\) Đặt \(x^2+2x=a\), pt trở thành:

\(a^2-3a+2=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=0\left(1\right)\\x^2+2x-2=0\left(2\right)\end{matrix}\right.\)

\(\left[{}\begin{matrix}\Delta\left(1\right)=4+4=8\\\Delta\left(2\right)=4+8=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{8}}{2}\\x=\dfrac{-2+\sqrt{8}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{12}}{2}\\x=\dfrac{-2+\sqrt{12}}{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{2}\\x=-1+\sqrt{2}\\x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{matrix}\right.\)

\(b,\) Đặt \(x^2+x=b\), pt trở thành:

\(b\left(b+1\right)-6=0\\ \Leftrightarrow b^2+b-6=0\\ \Leftrightarrow\left[{}\begin{matrix}b=2\\b=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\x\in\varnothing\left[x^2+x+3=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

\(d,x^4-2x^3+x=2\\ \Leftrightarrow x^4-2x^3+x-2=0\\\Leftrightarrow\left(x^3+1\right)\left(x-2\right)=0 \\ \Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x\in\varnothing\left[x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Lời giải:

a. 

PT $\Leftrightarrow (x^2+2x)^2-(x^2+2x)-2[(x^2+2x)-1]=0$

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

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

$\Leftrightarrow x^2+2x-1=0$ hoặc $x^2+2x-2=0$

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

b.

PT $\Leftrightarrow (x^2+x)^2+(x^2+x)-6=0$

$\Leftrightarrow (x^2+x)^2-2(x^2+x)+3(x^2+x)-6=0$

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

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

$\Leftrightarrow x^2+x-2=0$ (chọn) hoặc $x^2+x+3=0$ (loại do $x^2+x+3=(x+0,5)^2+2,75>0$)

$\Leftrightarrow x=-1\pm \sqrt{3}$

c. Nghiệm khá xấu. Bạn coi lại đề.

d.

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

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

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

$\Leftrightarrow x=-1$ hoặc $x=2$

d: Ta có: \(x^4-2x^3+x=2\)

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

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

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

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

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\) 

\(8 - x + 15 = 6 - 4x\)

\( - x + 4x = 6 - 8 - 15\)

\(3x =  - 17\)

\(x = \left( { - 17} \right):3\)

\(x = \dfrac{{ - 17}}{3}\)

Vậy nghiệm của phương trình là \(x = \dfrac{{ - 17}}{3}\).

b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\)

\( - 9 + 12u =  - 45 + 6u\)

\(12u - 6u =  - 45 + 9\)

\(u = \left( { - 36} \right):6\)

\(6u =  - 36\)

\(u =  - 6\)

Vậy nghiệm của phương trình là \(u =  - 6\).

c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\)

\(\left( {{x^2} + 6x + 9} \right) - \left( {{x^2} + 4x} \right) = 13\)

\({x^2} + 6x + 9 - {x^2} - 4x = 13\)

\(\left( {{x^2} - {x^2}} \right) + \left( {6x - 4x} \right) = 13 - 9\)

\(2x = 4\)

\(x = 4:2\)

\(x = 2\)

Vậy nghiệm của phương trình là \(x = 2\).

d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\)

\(\left( {{y^2} - 25} \right) - \left( {{y^2} - 4y + 4} \right) = 5\)

\({y^2} - 25 - {y^2} + 4y - 4 = 5\)

\(\left( {{y^2} - {y^2}} \right) + 4y = 5 + 4 + 25\)

\(4y = 34\)

\(y = 34:4\)

\(y = \dfrac{{17}}{2}\)

Vậy nghiệm của phương trình là \(y = \dfrac{{17}}{2}\).

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

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

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

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

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

a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)

b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)

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

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

\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)

c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)

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

\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)

c.

\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)

\(\Leftrightarrow2cos\left(x+12^0\right)=1\)

\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)

2.

Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:

\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)

\(\Rightarrow-1\le m\le\dfrac{1}{2}\)

a.

\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

b.

\(2x-10^0=arccot\left(4\right)+k180^0\)

\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)

2.

Phương trình \(sin\left(3x-27^o\right)=2m^2+m\) có nghiệm khi:

\(2m^2+m\in\left[-1;1\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m\le1\\2m^2+m\ge-1\end{matrix}\right.\)

\(\Leftrightarrow\left(m+1\right)\left(2m-1\right)\le0\)

\(\Leftrightarrow-1\le m\le\dfrac{1}{2}\)