-3*(x - 7) - 2(3x + 1) = 5x - 121

=> -3x + 21 - 6x - 2 - 5x + 121 = 0

=> -14x + 140 = 0

=> -14x = -140

=> x = 10

Vậy x = 10

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)

a) \(\left(2x+3\right)^2=\frac{9}{121}\)

Ta có: \(\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)

\(\Rightarrow2x+3\in\left\{\frac{3}{11};\frac{-3}{11}\right\}\)

\(\Rightarrow x\in\left\{\frac{-15}{11};\frac{-18}{11}\right\}\)

Vậy \(x\in\left\{\frac{-15}{11};\frac{-18}{11}\right\}\)

b) \(\left(3x-1\right)^3=\frac{-8}{27}\)

Ta có: \(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

\(\Rightarrow3x-1=\frac{-2}{3}\)

\(\Rightarrow x=\frac{1}{9}\)

Vậy \(x=\frac{1}{9}\)

a.

\(\left(2x+3\right)^2=\frac{9}{121}\)

\(\left(2x+3\right)^2=\left(\pm\frac{3}{11}\right)^2\)

\(2x+3=\pm\frac{3}{11}\)

TH1:

\(2x+3=\frac{3}{11}\)

\(2x=\frac{3}{11}-3\)

\(2x=-\frac{30}{11}\)

\(x=-\frac{30}{11}\div2\)

\(x=-\frac{15}{11}\)

TH2:

\(2x+3=-\frac{3}{11}\)

\(2x=-\frac{3}{11}-3\)

\(2x=-\frac{36}{11}\)

\(x=-\frac{36}{11}\div2\)

\(x=-\frac{18}{11}\)

Vậy \(x=-\frac{15}{11}\) hoặc \(x=-\frac{18}{11}\)

b.

\(\left(3x-1\right)^3=-\frac{8}{27}\)

\(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)

\(3x-1=-\frac{2}{3}\)

\(3x=-\frac{2}{3}+1\)

\(3x=\frac{1}{3}\)

\(x=\frac{1}{3}\div3\)

\(x=\frac{1}{9}\)

Chúc bạn học tốt ^^

e: =>-40+3+33+40-x=47

=>36-x=47

=>x=-11

f: =>x(x-3)(11-x)(11+x)=0

hay \(x\in\left\{0;3;11;-11\right\}\)

g: =>-62-38-x+2x=-100

=>x-100=-100

hay x=0

i: =>x-12-2x-31=6

=>-x-43=6

=>x+43=-6

hay x=-49

h: =>(x+1)=0

=>x=-1

f: =>x(x-3)(x+11)(x-11)=0

hay \(x\in\left\{0;3;-11;11\right\}\)

\(\left(2x+3\right)^2=\frac{9}{121}\)

\(\left(2x+3\right)^2=\left(\frac{3}{11}\right)^2\)

\(\Rightarrow2x+3=\frac{3}{11}\text{ hoặc }2x+3=\frac{-3}{11}\)

\(2x=\frac{3}{11}-3\text{ hoặc }2x=\frac{-3}{11}-3\)

\(2x=\frac{3}{11}-\frac{33}{11}\text{ hoặc }2x=\frac{-3}{11}-\frac{33}{11}\)

\(2x=\frac{-30}{11}\text{ hoặc }2x=\frac{-36}{11}\)

\(x=\frac{-30}{11}:2\text{ hoặc }x=\frac{-36}{11}:2\)

\(x=\frac{-30}{11}.\frac{1}{2}\text{ hoặc }x=\frac{-36}{11}.\frac{1}{2}\)

\(x=\frac{-15}{11}\text{ hoặc }x=\frac{-18}{11}\)

\(\left(x^2-3x+1\right)\left(21+3x-x^2\right)=121\)

\(\Leftrightarrow\left(x^2-3x+1\right)\left[22-\left(x^2-3x+1\right)\right]=121\)

Đặt \(x^2-3x+1=a\)

\(\Rightarrow a\left(22-a\right)=121\)

\(\Leftrightarrow a^2-22a+121=0\)

\(\Leftrightarrow\left(a-11\right)^2=0\)

\(\Leftrightarrow a=11\)

\(\Rightarrow x^2-3x+1=11\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{-2;5\right\}\)

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a) (x² - 121) . (2x + 3) = 0

=>x²-121=0 hoặc 2x+3=0

+)Nếu x²-121=0

=>x²=0+121=121

=>x²=(-11)² hoặc x²=11²

=>x=-11 hoặc x=11

+)Nếu 2x+3=0

=>2x=0-3=-3

=>x=(-3):2=\(\frac{-3}{2}\)

Vậy x=-11 hoặc x=11 hoặc x=\(\frac{-3}{2}\)
b) 2x² - 8x = 0

=>2x(x-4)=0

=>x=0 hoặc x-4=0

Nếu x-4=0

=>x=0+4=4

Vậy x=0 hoặc x=4
c) (3x + 1)⁵ = (3x + 1)⁴

=>(3x+1)⁵:(3x+1)⁴=(3x+1)⁴:(3x+1)⁴

=>3x+1=1

=>3x=1-1=0

=>x=0:3=0

Vậy x=0

a)(x² - 121) . (2x + 3) = 0

=>x²-121=0 hoặc 2x+3=0

• Với x²-121=0

<=>x²=121 <=>x=±11

• Với 2x+3=0

<=>2x=-3 <=>x=-3/2

b) 2x² - 8x = 0

=>2x(x-4)=0

=>2x=0 hoặc x-4=0

=>x=0 hoặc x=4