$x^2 \dfrac{1}{x^2}=x \dfrac{1}{x}$ $x\left(x 1\right)\left(x 2\right)\left(x 3\right)=24$ e,$\left(x^2 x 1\right)^2-2x^2-2x=5$ Giải pt

Bt: Giải các pt sau: a, $\dfrac{x+3}{x+1}$ + $\dfrac{x-2}{x}$=$\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}$ b, $\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1$ c, $\dfrac{1}{x^2+x+1}-\dfrac{1}{x^2-x+1}=\dfrac{1-2x}{x^4+x^2+1}$ d, $\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1$ e,...

1

ZH

ミ★Zero ❄ ( Hoàng Nhật )

12 tháng 3 2023

$a.x^2+\dfrac{1}{x^2}=x+\dfrac{1}{x}$ ( ĐKXĐ : $x\ne0$ )

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

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

$\Leftrightarrow\left[{}\begin{matrix}1-x=0\\\dfrac{1}{x^2}-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\1-x^3=0\end{matrix}\right.$

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

$\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\Leftrightarrow x=1$ ( x² + x + 1 loại nhé nếu phân tích ra thì ta được $x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R$ )

Vậy $S=\left\{1\right\}$

b, $x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24$

$\Leftrightarrow x\left(x+3\right).\left(x+1\right)\left(x+2\right)-24=0$

$\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0$

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

$\Leftrightarrow\left(x^2+3x+1\right)-1-24=0\Leftrightarrow\left(x^2+3x+1\right)-25=0$

$\Leftrightarrow\left(x^2+3x+1-5\right)\left(x^2+3x+1+5\right)=0\Leftrightarrow\left(x^2+3x-4\right)\left(x^2+3x+6\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+4\right)=0\\\left(x+\dfrac{3}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\end{matrix}\right.$

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

Vậy $S=\left\{-4;1\right\}$

e, $\left(x^2+x+1\right)-2x^2-2x=5\Leftrightarrow\left(x^2+x+1\right)-2x^2-2x-2-3=0$

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

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

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

$\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0$ ( x^2 + x + 2 loại nhé y như mấy câu trên luôn khác 0 ! )

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

Vậy $S=\left\{-2;1\right\}$

Đúng(2)

GP

Giang Phạm

23 tháng 2 2019

1

AL

Ánh Lê

23 tháng 2 2019

a) Đk : $x\ne0;\ne1$

$\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}$

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

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

$\Rightarrow\dfrac{0}{x-1}=0$

=> Phương trình có vô số nghiệm x

b) Đk : $x\ne2;x\ne3$

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

$\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}$

=0

$\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0$

$\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0$

=> Phương trình vô nghiệm

c)

$\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0$

$\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0$

$\Rightarrow\dfrac{-1}{x^4+x^2+1}=0$

=> PTVN

d) Thôi tự làm đi, câu này dễ :Vvv

e)

$\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)$=40

$\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40$

$\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40$

Đặt

$x^2+6x+7=t$

Phương trình tương đương

$\left(t-1\right)\left(t+1\right)=40$

$t^2=41$

$t=\pm\sqrt{41}$

Thay vào tìm x.

GP

Giang Phạm

24 tháng 2 2019

Thanks ;)

NN

nguyen ngoc son

16 tháng 2 2021

giải phương trình

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

b.$x\left(2x-9\right)=3x\left(x-5\right)$

c.$3x-15=2x\left(x-5\right)$

d.$\dfrac{5-x}{2}=\dfrac{3x-4}{6}$

e.$\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}$

1

NL

Nguyễn Lê Phước Thịnh

16 tháng 2 2021

a) 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: $S=\left\{\dfrac{3}{2};4\right\}$

b) Ta có: $x\left(2x-9\right)=3x\left(x-5\right)$

$\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0$

$\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0$

$\Leftrightarrow x\left(2x-9-3x+15\right)=0$

$\Leftrightarrow x\left(6-x\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.$

Vậy: S={0;6}

c) Ta có: $3x-15=2x\left(x-5\right)$

$\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0$

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

$\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.$

Vậy: $S=\left\{5;\dfrac{3}{2}\right\}$

d) Ta có: $\dfrac{5-x}{2}=\dfrac{3x-4}{6}$

$\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)$

$\Leftrightarrow30-6x=6x-8$

$\Leftrightarrow30-6x-6x+8=0$

$\Leftrightarrow-12x+38=0$

$\Leftrightarrow-12x=-38$

$\Leftrightarrow x=\dfrac{19}{6}$

Vậy: $S=\left\{\dfrac{19}{6}\right\}$

e) Ta có: $\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}$

$\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}$

$\Leftrightarrow6x+4-3x-1=12x+10$

$\Leftrightarrow3x+3-12x-10=0$

$\Leftrightarrow-9x-7=0$

$\Leftrightarrow-9x=7$

$\Leftrightarrow x=-\dfrac{7}{9}$

Vậy: $S=\left\{-\dfrac{7}{9}\right\}$

Đúng(2)

HH

Hoàng Hà Tiên

9 tháng 2 2021

a$8\left(x+\dfrac{1}{x}\right)^{2^{ }}+4\left(x^{2^{ }}+\dfrac{1}{x^2}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)=\left(x+4\right)^2$giải các phương trình$\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}$

1

NL

Nguyễn Lê Phước Thịnh

9 tháng 2 2021

b)

ĐKXĐ: $x\notin\left\{2;3;\dfrac{1}{2}\right\}$

Ta có: $\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}$

$\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}$

$\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}$

Suy ra: $x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10$

$\Leftrightarrow2x^2-14=2x^2+x-10$

$\Leftrightarrow2x^2-14-2x^2-x+10=0$

$\Leftrightarrow-x-4=0$

$\Leftrightarrow-x=4$

hay x=-4(nhận)

Vậy: S={-4}

giải các PT sau :a) $\left|2x+3\right|-\left|x\right|+\left|x-1\right|=2x+4$b) $\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0$c) $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}$d) $x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=4$e) $\sqrt{4x+3}+\sqrt{2x+1}=6x+\sqrt{8x^2+10x+3}-16$f)$\sqrt[3]{x-2}+\sqrt{x+1}=3$GIÚP MÌNH VỚI MÌNH ĐANG CẦN...

LT

Lê Thu Trang

6 tháng 12 2021

0

K

Kitana

24 tháng 4 2021

Giải pt sau

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

1

NV

Nguyễn Việt Lâm

Giáo viên

24 tháng 4 2021

ĐKXĐ: ...

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

Đặt $\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.$

$\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.$

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

$\Leftrightarrow...$

Giải bpt sau a, $\left(x+3\right)^2-\left(x-3\right)^2\le3\left(x+1 \right)$ b, $2\left(x+3\right).\left(x+4\right)\left(x-2\right)^2+\left(x-1\right)^2$ c, $5x^2-18x+19-\left(2x-3\right)^20$ d, $\dfrac{\left(3x-2\right)^2}{4}-\dfrac{3\left(x-2\right)}{8}-1\dfrac{-15x\left(5-3x\right)}{2}$ e, $2x^2+2x+2-\dfrac{15\left(x-1\right)}{2}-12x\left(x-2,75\right)$ g, \(\dfrac{5x^2-3}{5}+\dfrac{3x-1}{4}...

BT

Bành Thụy Hóii

30 tháng 3 2018

3

ND

Nhã Doanh

30 tháng 3 2018

Hỏi đáp Toán

PT

Phương Trâm

30 tháng 3 2018

Dài quá c ơi :<

LS

Lê Song Phương

26 tháng 2 2023

Giải các pt

a) $\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)=3\sin x+\cos x+2$

b) $\dfrac{\left(2-\sqrt{3}\right)\cos x-2\sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2\cos x-1}=1$

c) $2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x\right)\sin x=1$

3

NV

Nguyễn Việt Lâm

Giáo viên

26 tháng 2 2023

a.

$\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2$

$\Leftrightarrow sin2x+cos2x=3sinx+cosx+2$

$\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0$

$\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0$

$\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.$

$\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1$

$\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}$

$\Leftrightarrow...$

NV

Nguyễn Việt Lâm

Giáo viên

26 tháng 2 2023

b.

ĐKXĐ: $cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.$

$\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1$

$\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx$

$\Leftrightarrow-\sqrt{3}cosx+sinx=0$

$\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0$

$\Rightarrow x-\dfrac{\pi}{3}=k\pi$

$\Rightarrow x=\dfrac{\pi}{3}+k\pi$

Kết hợp ĐKXĐ $\Rightarrow x=\dfrac{4\pi}{3}+k2\pi$

Bài 1. Giải các bất phương trình sau 1) $\dfrac{2x-1}{x+1}-2< 0$ 2) $\dfrac{x^2-2x+5}{x-2}-x+1\ge0$3) $\dfrac{\left(1+2x\right)\left(x-3\right)}{\left(2x+3\right)\left(1-x\right)}\le0$ 4) $\left|2x-3\right|>5$ ...

FA

FREESHIP Asistant

28 tháng 1 2022

2

TH

Thanh Hoàng Thanh

28 tháng 1 2022

$\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.$

Mà $-3< 0.$

$\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).$

$\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.$

$\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.$

Kết hợp ĐKXĐ.

$\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.$

$\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).$

Đặt $\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).$

Ta có bảng sau:

$x$	$-\infty$ $-\dfrac{3}{2}$ $-\dfrac{1}{2}$ $1$ $2$ $+\infty$
$1+2x$	- \| - 0 + \| + \| +
$x-2$	- \| - \| - \| - 0 +
$2x+3$	- 0 + \| + \| + \| +
$1-x$	+ \| + \| + 0 - \| -
$f\left(x\right)$	- \|\| + 0 - \|\| + 0 -

Vậy $f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup$(1;2].

MS

Mot So

28 tháng 1 2022

2) $Đ K : x \neq 2$

Nếu $x > 2$

BPT ⇔ $x^{2} - 2 x + 5 - (x - 1) (x - 2) \geq 0$ ⇔ $x^{2} - 2 x + 5 - (x^{2} - 3 x + 3) \geq 0$

⇔ $x + 2 \geq 0$ ⇔ $x \geq - 2$ ⇒ Lấy $x \geq 2$

Nếu $x < 2$

$x < 2$ $\frac{- (x^{2} - 2 x + 5)}{x - 2} - x + 1 \geq 0$ ⇔ $- x^{2} + 2 x - 5 - (x - 1) (x - 2) \geq 0$

Giải các bất phương trình saua/ (x+1).(x-1).(3x-6)>0b/ $\dfrac{x+3}{x-2}\le0$c/ $\dfrac{\left(2x-5\right).\left(x+2\right)}{-4x+3}\ge0$d/ \(\dfrac{2x-5}{3x+2}...

HP

Heo Peppa

18 tháng 9 2023