\(a,\left(x-3\right)^2-4=0\)

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

\(\Rightarrow x-3=\pm2\)

\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)

Vậy \(x=5\)hoặc \(x=1\)

\(b,x^2-2x=24\)

\(\Leftrightarrow x^2-2x+1-1=24\)

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

\(\Leftrightarrow x-1=\pm5\)

\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)

Vậy \(x=6\) hoặc \(x=-4\)

\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)

\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow10x+255=0\)

\(\Leftrightarrow10x=-255\)

\(\Leftrightarrow x=\frac{-51}{2}\)

\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x-27=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

\(e,\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)

\(\Leftrightarrow9x^2-6x+1+2\left(x^2+6x+9\right)+11\left(x-x^2+x-x\right)=6\)

\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18-11x^2+11x=6\)

\(\Leftrightarrow17x+19=6\)

\(\Leftrightarrow17x=-13\)

\(\Leftrightarrow x=\frac{-13}{17}\)

a: \(\Leftrightarrow2x-2+4x+8=-12\)

=>6x+6=-12

=>6x=-18

hay x=-3

b: \(\Leftrightarrow-10x-15-12+9x=13\)

=>-x-27=13

=>-x=40

hay x=-40

c: \(\Leftrightarrow-10x+70+20-5x=-15\)

\(\Leftrightarrow-15x=-105\)

hay x=7

d: \(\Leftrightarrow8x-12-7x+14=10\)

=>x+2=10

hay x=8

e: \(\Leftrightarrow-12x-18+14x+2=2\)

=>2x-16=2

hay x=9

a) (x + 6)(3x + 1) + x² - 36 = 0

<=> 3x² + x + 18x + 6 + x² - 36 = 0

<=> 4x² + 19x - 30 = 0

<=> 4x² + 24x - 5x - 30 = 0

<=> 4x(x + 6) - 5(x + 6) = 0

<=> (x + 6)(4x - 5) = 0

<=> x + 6 = 0 hoặc 4x - 5 = 0

<=> x = -6 hoặc x = 5/4

Bài 1 mình đã làm xong rồi, anh em nào giúp mình bài 2 với!

a/ | x + 10 | = 15

=> x + 10 = 15 hay x + 10 = - 15

+/ x + 10 = 15

=> x = 15 - 10 = 5

+/ x + 10 = - 15

=> x = -15 - 10 = -25

Vậy x thuộc {5; - 25}

b/ | x - 3 | + 5 = 7

=> | x - 3 | = 7 - 5 = 2

=> x - 3 = 2 hay x - 3 = -2

+/ x - 3 = 2

=> x = 2+3 = 5

+/ x - 3 = -2

=> x = -2 + 3 = 1

Vậy x thuộc {5;1}

c/ | x - 3 | + 12 = 6

=> | x - 3 | = 6 - 12 = - 6

Vì | x - 3 | luôn > 0

mà | x - 3 | = - 6

Vậy k có giá trị của x

d/ (2x + 4) . (3x + 9) = 0

=> 2x + 4 = 0 hoặc 3x + 9 = 0

+/ 2x + 4 = 0

=> 2x = 0 - 4 = -4

=> x = (-4) / 2 = -2

+/ 3x - 9 = 0

=> 3x = 0 + 9 = 9

=> x = 9 / 3 = 3

Vậy x thuộc {-2;3}

a. \(\left|x+10\right|=15\)

\(\Rightarrow x+10=\pm15\)

\(TH1:x+10=15\)

\(x=15-10\)=5

TH2: x + 10 = -15

x = -15 -10 = -25

Vậy x \(\in\left\{5;-25\right\}\)

b. \(\left|x-3\right|+5=7\)

\(\left|x-3\right|=7-5=2\)

\(\Rightarrow x-3=\pm2\)

TH1: x - 3 = 2

x = 2 + 3 = 5

TH2: x - 3 = -2

x = -2 + 3 = 1

Vậy x \(\in\left\{5;-1\right\}\)

* Đối với bài tập về phép đối này thì có 2 trường hợp, giải TH âm và dương của số đã cho bên kết quả.

Mỏi tay, xl

bài e đề là x² - 5x = 0 ak

a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)

hay x=0

Vậy: x=0

b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)

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

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

Vậy: \(x=-\dfrac{1}{4}\)

c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)

hay \(x=\dfrac{-1}{9}\)

Vậy: \(x=\dfrac{-1}{9}\)

\(a.x=\dfrac{1}{3}-\dfrac{1}{3}\)

\(x=0\)

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

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

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

c. \(\dfrac{-1}{6}=\dfrac{3}{2x}\)

\(-2x=18\)

\(x=-9\)

d. \(\dfrac{4}{5}=\dfrac{-12}{9-x}\)

\(4.\left(9-x\right)=-60\)

\(9-x=-15\)

\(x=24\)

\(e.\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\left(x+1\right)^2=9\)

\(\left[{}\begin{matrix}x+1=-3\\x+1=3\end{matrix}\right.\)

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

f.\(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)

\(\left(x-1\right)^2=16\)

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

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

bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((

Ukm

It's very hard

l can't do it 

Sorry!

a) \(x^4-x^3-7x^2+x+6=0\)

\(\Leftrightarrow x^4+2x^3-3x^3-6x^2-x^2-2x+3x+6=0\)

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

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

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

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

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

b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)

\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)

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

Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)

\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)

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

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

\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)

Từ đó tính đc x

d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)

\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

Đặt \(x^2+5x+5=a\), khi đó pt có dạng:

\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Bài 1 Tìm x biết:

a)65-(29-x)=32

65 -29+x=31

x=31-65+29

x=-5

b)(x+5)-(x+23)=x-34

x+5 -x +23 = x-34

(x-x)+ (23+5)=x-34

0+28=x-34

28=x-34

28+34=x

62=x

=>x=62

c)(16-x)+(x-38)=x+44

16-x+x-38=x+44

-x+x-x=44-16+38

-x=36

=>x=-36

d)-12+3(-x+7)=-18

3(-x+7)=-18+12

3(-x+7)=-6

-x+7=-6:3

-x+7=-2

-x=-2-7

-x=-9

=>x=9

Baif 2

d)|7-x|=10

=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)

e)(x-6).(7-2x)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)

f)(9-x).(2x+8)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

g)x(-x+8).(-3x-18)=0

\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)

h)(-x+8).(x-54).(-24-x)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)