1) 2x - 378 = 122

=> 2x = 122 + 378

=> 2x = 500

=> x = 500 : 2

=> x = 250

2) 8x - 4x = 1208

=> 4x = 1208

=> x = 1208 : 4

=> x = 302

3) x - 382 = 159 : 3

=> x - 382 = 53

=>  x = 53 + 382

=> x = 435

4) 3x + 30 = 420

=> 3x = 420 - 30

=> 3x = 390

=> x = 390 : 3

=> x = 130

5) 12x  - 13x - 500 = 1500

=> -x = 1500 + 500

=> -x = 2000

=> x = -2000

1,\(2x-378=122\)

\(2x=122+378\)

\(2x=500\)

\(x=500:2\)

\(x=250\)

2,\(8x-4x=1208\)

\(x.\left(8-4\right)=1208\)

\(x.4=1208\)

\(x=1208:4\)

\(x=302\)

3,\(x-382=159:3\)

\(x-382=53\)

\(x=52+382\)

\(x=434\)

4,\(3x+30=420\)

\(3x=420-30\)

\(3x=390\)

\(x=390:3\)

\(x=130\)

5,\(12x-13x-500=1500\)

như câu 2 

\(1,2x-378=122\)\(\Rightarrow2x=500\Rightarrow x=250\)

\(2,8x-4x=1208\)\(\Rightarrow4x=1208\Rightarrow x=302\)

\(3,x-382=159\div3\)\(\Rightarrow x-382=53\Rightarrow x=435\)

\(4,3x+30=402\)\(\Rightarrow3x=372\Rightarrow x=124\)

\(5,12x-13x-500=1500\)\(\Rightarrow-x-500=1500\Rightarrow-x=1000\Leftrightarrow x=-1000\)

\(\left(x-5\right).4=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\4=0\left(ktm\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\4=0\left(ktm\right)\end{cases}}}\)

vậy x=5

\(8x-4x=1208\)

\(x\left(8-4\right)=1208\)

\(x4=1208\)

\(\Rightarrow x=1208:4=302\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)

vậy x=4 hoặc x=3

đề sai \(3^x:3^2=243\)

\(3^{x-2}=243\)

\(3^{x-2}=3^5\)

\(\Rightarrow x-2=5\)

\(\Rightarrow x=7\)

\(2^x.64=2^6\)

\(2^x.2^6=2^6\)

\(2^{x+6}=2^6\)

\(\Rightarrow x+6=6\)

\(\Rightarrow x=0\)

a) x= - 15

b) x= - 2

c) x= -12 

d) x= -2 

1.

\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)

\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)

\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max

2.

\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)

\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)

\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)

\(E_{min}=-1\) khi \(x=0\)

\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)

\(G_{min}=-2\) khi \(x=2\)

a)2x + 15 = 27

2x = 12

x = 6

b.1)8x - 4x = 1208

4x = 1208

x = 302

b.2)a : 3 = 10

=> a = 10 x 3

=> a = 30

Bài 1 :                                                         Bài giải

a, \(2x+15=27\)

\(2x=12\)

\(x=6\)

b, \(8x-4x=1208\)

\(4x=1208\)

\(x=302\)

Bài 2 :                                             Bài giải

                Gọi số cần tìm là a ( \(a\inℕ^∗\) )

Ta có : a : 3 = 10

       a = 30

Vậy số cần tìm là 30

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

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

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

A. \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+3x+2x+6\right)-\left(x^2+5x-2x-10\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow x^2+3x+2x-x^2-5x+2x=-6-10\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\) .Vậy \(S=\left\{-8\right\}\)

B. \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x+5x-20\)
\(\Leftrightarrow2x^2-8x+3x+x^2-2x-5x-3x^2+12x-5x=12-10-20\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\) . Vậy \(S=\left\{\dfrac{18}{5}\right\}\)

C. \(\left(8-4x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow8x-4x^2-8x+4x^2+4x-8x=-16+8\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\) . Vậy \(S=\left\{2\right\}\)

D. \(\left(2x-3\right)\left(8x+2\right)=\left(4x+1\right)\left(4x-1\right)-3\)
\(\Leftrightarrow16x^2+4x-24x-6=16x^2+1^2-3\)
\(\Leftrightarrow16x^2+4x-24x-16x^2=6+1-3\)
\(\Leftrightarrow-20x=4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\) . Vậy \(S=\left\{-\dfrac{1}{5}\right\}\)

a)(x+2)(x+3)-(x-2)(x+5)=0

\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)

<=>2x=-16

<=>x=-8

b)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow5x=22\Leftrightarrow x=\dfrac{22}{5}\)

c)(8-4x)(x+2)+4(x-2)(x+1)=0

\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)

\(\Leftrightarrow-4x=-8\Leftrightarrow x=2\)

d)(2x-3)(8x+2)=(4x+1)(4x-1)-3

\(\Leftrightarrow16x^2+4x-24x-6=16x^2-4x+4x-1-3\)

\(\Leftrightarrow-20x=-2\Leftrightarrow x=\dfrac{-1}{10}\)

a) Ta có: \(\dfrac{1}{x+3}+\dfrac{8-x}{4x^2+8x}\)

\(=\dfrac{1}{x+3}+\dfrac{8-x}{4x\left(x+2\right)}\)

\(=\dfrac{4x\left(x+2\right)}{4x\left(x+3\right)\left(x+2\right)}+\dfrac{\left(8-x\right)\left(x+3\right)}{4x\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{4x^2+8x+8x+24-x^2-3x}{4x\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{3x^2+13x+24}{4x\left(x+3\right)\left(x+2\right)}\)

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

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

\(=\dfrac{3x+15-2x^2-10x+x^2+2x-5x-10}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}\)

\(=\dfrac{-x^2-10x+5}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}\)