1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)

\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)

\(x\times\left(8-25\right)=2\times29-33\)

\(x\times-17=25\)

\(x=-\dfrac{25}{17}\)

2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)

\(15\div\left(x+2\right)=\left(27+3\right)\div1\)

\(15\div\left(x+2\right)=30\div1\)

\(15\div\left(x+2\right)=30\)

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

\(x=-\dfrac{3}{2}\)

3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)

\(20\div\left(x+1\right)=\left(25+1\right)\div13\)

\(20\div\left(x+1\right)=26\div13\)

\(20\div\left(x+1\right)=2\)

\(x+1=20\div2\)

\(x+1=10\)

\(x=9\)

4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)

\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)

\(320\div\left(x-1\right)=100\div4+15\)

\(320\div\left(x-1\right)=25+15\)

\(320\div\left(x-1\right)=40\)

\(x-1=8\)

\(x=9\)

5) \(240\div\left(x-5\right)=2^2\times5^2-20\)

\(240\div\left(x-5\right)=4\times25-20\)

\(240\div\left(x-5\right)=100-20\)

\(240\div\left(x-5\right)=80\)

\(x-5=30\)

\(x=35\)

6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)

\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)

\(70\div\left(x-3\right)=80\div4-10\)

\(70\div\left(x-3\right)=20-10\)

\(70\div\left(x-3\right)=10\)

\(x-3=7\)

\(x=10\)

a) Gọi x²=a 

=> 3a² - a - 234=0

∆=b² - 4ac= (-1)²-4×3×(-234)=2809

√∆=53

∆>0 nên pt có 2 nghiệm phân biệt 

a1=-b+√∆/2a = -(-1)+53/2×3 =9

a2=-b-√∆/2a = -(-1)-53/2×3 =-26/3

Thay x²=a=9 =>x=3,x=-3

          x²=a=-26/3 (loại)

Vậy nghiệm của pt là x =3, x=-3

d) (x+4)(x+5)(x+7)(x+8)=4

<=> (x+4)(x+8)(x+5)(x+7)=4

<=> (x²+8x+4x+32)(x²+7x+5x+35)=4

<=> (x²+12x+32)(x²+12x+35)=4

Đặt t=x²+12x+32

=> t(t+3)=4

<=> t²+3t-4=0

     (a=1,b=3,c=-4)

a+b+c=1+3+(-4)=0

=> t1=1 ; t2= c/a =-4/1=-4

Thay t=x²+12x+32=1

=> x²+12x+31=0

∆=b²-4ac= 12² -4×1×31= 20

√∆=2√5

∆>0 nên pt có 2 nghiệm phân biệt 

x1=-b+√∆/2a= -12+2√5/2×1= -6+√5

x2=-b-√∆/2a = -12-2√5/2×1= -6-√5

Thay t=x²+12x+32=-4

=> x²+12x+36=0

∆=b²-4ac= 12²-4×1×36=0

∆=0 nên pt có nghiệm kép 

x1=x2= -b/2a= -12/2×1 = -6

Vậy nghiệm của pt là S={-6+√5 ; -6-√5; -6}

b: =>\(\dfrac{-x}{x-1}=\dfrac{2x+3}{\left(x-1\right)\left(x+2\right)}\)

=>-x^2-2x-2x-3=0

=>x^2+4x+3=0

=>x=-1 hoặc x=-3

c: =>x^3+x^2-3x-3x^2+2x+2=0

=>x^3-2x^2-x+2=0

=>(x-2)(x-1)(x+1)=0

=>\(x\in\left\{2;1;-1\right\}\)

d: =>(x^2+12x+32)(x^2+12x+35)-4=0

=>(x^2+12x)^2+67(x^2+12x)+1116=0

=>(x^2+12x+36)(x^2+12x+31)=0

=>\(x\in\left\{-6;-6+\sqrt{5};-6-\sqrt{5}\right\}\)

544

8

200

a) x= 864- 320

x = 544

b) 3 x x = 24

x = 24:3

x = 8

c) x:5= 40

x = 40x5

x = 200

a)

x = 864- 320

x = 544

b) 3 \(\times\) x = 24

x = 24:3

x = 8

c) x : 5 = 40

    x  = 40 : 5

    x  =  8

chọn câu trả lời của mk nha!!!

a)

4 x 9 = 9 x 4

b)

5 x 10 = 10 x 5

c)

3 112 x 8 = 8 x 3 112

d)

41 320 x 3 = 3 x 41 320

4(x-3)²-320 = 0

=> 4(x-3)² = 320

=> (x-3)² = 320 : 4 = 80 = (8,94427191)² = (-8,94427191)²

TH1:

x - 3 = 8,94427191

=> x = 11,94427191

TH2: 

x - 3 = -8,94427191

=> x = -5,94427191

7(4+x)³-875 = 0

=> 7(4+x)³ = 875

=> (4+x)³ = 875:7 = 125 = 5³

=> 4 + x = 5

=> x = 1

650 - 5(x+4)² = 330

5(x+4)² = 650 - 330 =320

=> (x+4)² = 320 : 5 = 64 = 8²

=> x+4 = 8

=> x = 4

3(5-x)²-15 = 60

=> 3(5-x)² = 75

=> (5-x)² = 25 = 5² =(-5)²

TH1:

5-x =5

=> x = 0

TH2: 5-x = -5

=> x = 10

a) 234 x 95 + 234 x 4 + 234 = 234 x ( 95 + 4 + 1 ) = 234 x 100 = 23400

a).234 x 95 + 234 x 4 + 234 

= 234 x ( 95 + 4 + 1 ) 

= 234 x 100

= 23400

b)  ( 1+1/2) x ( 1+1/3) x ( 1+1/4 )

= ( 2/2+1/2 ) x ( 3/3+1/3 ) x ( 4/4 + 1/4)

=3/2 + 4/3 + 5/4

= 18/12 + 16/12 + 20/12

= 54/12

= 9 /2

b) ( 1 + \(\frac{1}{2}\)) x ( 1 + \(\frac{1}{3}\) ) x ( 1 + \(\frac{1}{4}\) ) = \(\frac{3}{2}\) x \(\frac{4}{3}\) x \(\frac{5}{4}\) = \(\frac{5}{2}\)

a) 3^x.9^6=3^14              b,27^9.(3^x) = 3^31

=3^x.3^12=3^14                =>3^27.(3^x)^2=3^31

=> 3^x+12=3^14                => (3^x)^2=3^31.3^27

=> x=2                               =>(3^x)^2=3^4=>2^x=4

                                          =>x=2

a, \(\left(x+1\right)^2=169\)

\(\left(x+1\right)^2=13^2\)

\(x+1=13\)

\(x=13-1\)

\(x=12\)

1.

a) \(\left(x+1\right)^2=169\)

⇒ \(x+1=\pm13\)

⇒ \(\left[{}\begin{matrix}x+1=13\\x+1=-13\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=13-1\\x=\left(-13\right)-1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=12\\x=-14\end{matrix}\right.\)

Vậy \(x\in\left\{12;-14\right\}.\)

b) \(\left(x+3\right)^3=-\frac{1}{27}\)

⇒ \(\left(x+3\right)^3=\left(-\frac{1}{3}\right)^3\)

⇒ \(x+3=-\frac{1}{3}\)

⇒ \(x=\left(-\frac{1}{3}\right)-3\)

⇒ \(x=-\frac{10}{3}\)

Vậy \(x=-\frac{10}{3}.\)

c) \(\left(2x-4\right)^4=\frac{1}{625}\)

⇒ \(2x-4=\pm\frac{1}{5}\)

⇒ \(\left[{}\begin{matrix}2x-4=\frac{1}{5}\\2x-4=-\frac{1}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=\frac{1}{5}+4=\frac{21}{5}\\2x=\left(-\frac{1}{5}\right)+4=\frac{19}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{21}{5}:2\\x=\frac{19}{5}:2\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=\frac{21}{10}\\x=\frac{19}{10}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{21}{10};\frac{19}{10}\right\}.\)

Còn câu d) bạn làm tương tự như mấy câu trên.

Chúc bạn học tốt!