giải 3²x+3x=121¹-1¹²¹
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-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 \(x\in\left\{0;3;-11;11\right\}\)
\(\left(x^2-3x+1\right)\left(21+3x-x^2\right)=121\)
\(\Leftrightarrow\left(x^2-3x+1\right)\left[22-\left(x^2-3x+1\right)\right]=121\)
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 \(S=\left\{-2;5\right\}\)
a ) Đặt t = x - 7 .
=> pt trở thành : \(\left(t-1\right)^4+\left(t+1\right)^4=16\)
b ) Phân tích :
\(x^3-x^2+2x^2-2x+3x-3=0\)
NHhóm rồi sẽ ra
\(\left(2x+3\right)^2=\frac{9}{121}\)
\(\left(2x+3\right)^2=\left(\frac{3}{11}\right)^2\)
\(\Rightarrow2x+3=\frac{3}{11}\text{ hoặc }2x+3=\frac{-3}{11}\)
\(2x=\frac{3}{11}-3\text{ hoặc }2x=\frac{-3}{11}-3\)
\(2x=\frac{3}{11}-\frac{33}{11}\text{ hoặc }2x=\frac{-3}{11}-\frac{33}{11}\)
\(2x=\frac{-30}{11}\text{ hoặc }2x=\frac{-36}{11}\)
\(x=\frac{-30}{11}:2\text{ hoặc }x=\frac{-36}{11}:2\)
\(x=\frac{-30}{11}.\frac{1}{2}\text{ hoặc }x=\frac{-36}{11}.\frac{1}{2}\)
\(x=\frac{-15}{11}\text{ hoặc }x=\frac{-18}{11}\)
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
<=>x2=121 <=>x=±11
<=>2x=-3 <=>x=-3/2
b) 2x2 - 8x = 0
=>2x(x-4)=0
=>2x=0 hoặc x-4=0
=>x=0 hoặc x=4
\(3^2x+3x=121-1\\ =>9x+3x=120\\ =>12x=120\\ =>x=120:12\\ =>x=10\)
32x+3x=1211-1121
=> 9x+3x=121-1
=> 12x=120
=> x=10