Tìm x biết : \(\left(x-1\right):0,16=-9:\left(1-x\right)\)
Những câu hỏi liên quan
Tìm x:
1) \(\text{(x−1):0,16=−9:(1−x)}\)
2) \(\left(\left|x\right|-\dfrac{3}{2}\right)\left(2x^2-10\right)=0\)
3)\(8\sqrt{x}=x^2\left(x\ge0\right)\)
1) (x−1):0,16=−9:(1−x)
\(\Rightarrow\)(x-1):0,16= 9:(-1):(x-1)
\(\Rightarrow\)(x-1):0,16=9:(x-1)
\(\Rightarrow\)(x-1).(x-1)= 9. 0,16
\(\Rightarrow\)(x-1)\(^2\)= 1,44=1,2\(^2\)=(-1,2)\(^2\)
\(\Rightarrow\)x-1=1,2\(\Rightarrow\)x=2,2
hoặc x-1= -1,2\(\Rightarrow\)x= -0,2
Vậy x =2,2 ; x=0,2
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Tìm x, y biết:
a) \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
b) (3-x) : 0,16 = -9 : (x-3)
c) \(\left(3x-\frac{4}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2=0\)
d) \(\left(x+y-\frac{1}{2}\right)^2+\left(x-y+\frac{1}{6}\right)^2=0\)
e) x. \(\left(x-\frac{1}{4}\right)\)> 0
Tìm x biết : \(\left(\frac{1}{3}\right)^x\left(\frac{1}{9}\right)^x\left(\frac{1}{27}\right)^x\left(\frac{1}{81}\right)^x\left(\frac{1}{243}\right)^x=\left(-\frac{1}{3}\right)^{30}\)
TÌM X, BIẾT
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x^2-x-6x-6\)
\(\Leftrightarrow18x+16=x^2-7x-6\)
\(\Leftrightarrow x^2-7x-18x=16+6\)
\(\Leftrightarrow x^2-15x=22\)
\(\Leftrightarrow x^2-15x-22=0\)
......
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Tìm x biết:
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)
\(\Leftrightarrow\left(6x^2+27x+4x+18\right)-\left(6x^2+x+12x+2\right)=x-1-x+6\)
\(\Leftrightarrow6x^2+31x+18-6x^2-x-12x-2=7\)
\(\Leftrightarrow18x+16=7\)
\(\Leftrightarrow18x=-9\)
\(\Leftrightarrow x=\frac{-1}{2}\)
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\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)
\(3x\left(2x+9\right)+2\left(2x+9\right)-x\left(6x+1\right)-2\left(6x+1\right)=x-1-x+6\)
\(6x^2+27x+4x+18-6x^2-x-12x-2=5\)
\(6x^2+\left(27x+4x\right)+18-6x^2-\left(12x+x\right)-2=5\)
\(6x^2+31x+18-6x^2-13x-2=5\)
\(\left(6x^2-6x^2\right)+\left(31x-13x\right)+\left(18-2\right)=5\)
\(18x+16=5\)
\(18x=5+16\)
\(18x=21\)
\(x=21:18\)
\(x=\frac{7}{6}\)
Vậy \(x=\frac{7}{6}\)
P/s: Mình mới lớp 6 nên hi vọng bn xem bài của mik thật kĩ xem có sai sót không,cảm ơn.
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9 tháng 8 2023 lúc 20:51
Tìm x biết:
\(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=2
=>x^3-3x^2+3x-1-x^3-27+3x^2-12=2
=>3x-40=2
=>x=42/3=14
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tìm x biết a)x+2x+3x+4x+...+2015x2016times2017b)1-3+3^2-3^3+...+left(-3right)^xfrac{9^{1008}-1}{4}c)left|x+1right|+left|x+2right|+...+left|x+100right|605xd)tìm x nguyên biết left|x-1right|+left|x-2right|+...+left|x-100right|2500e) tìm x nguyên biết 2004left|x-4right|+left|x-10right|+left|x+101right|+left|x+99xright|+left|x+1000right|
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tìm x biết
a)\(x+2x+3x+4x+...+2015x=2016\times2017\)
b)\(1-3+3^2-3^3+...+\left(-3\right)^x=\frac{9^{1008}-1}{4}\)
c)\(\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=605x\)
d)tìm x nguyên biết \(\left|x-1\right|+\left|x-2\right|+...+\left|x-100\right|=2500\)
e) tìm x nguyên biết \(2004=\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|x+99x\right|+\left|x+1000\right|\)
a) Tìm tập hợp các số nguyên x, biết rằng4dfrac{5}{9}:2dfrac{5}{18}-7 x left(3dfrac{1}{5}:3,2+4,5.1dfrac{31}{45}right):left(-21dfrac{1}{2}right)b) tìm x, biết left|x+dfrac{1}{2}right|+left|x+dfrac{1}{6}right|+left|x+dfrac{1}{12}right|+left|x+dfrac{1}{20}right|+....+left|x+dfrac{1}{110}right|-11xc)Tính gt biểu thức C2x^3-5y^3+2015 tại x,y thỏa mãn left|x-1right|+left(y+2right)^{20}0
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a) Tìm tập hợp các số nguyên x, biết rằng\(4\dfrac{5}{9}:2\dfrac{5}{18}-7< x< \left(3\dfrac{1}{5}:3,2+4,5.1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\)
b) tìm x, biết \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+\left|x+\dfrac{1}{12}\right|+\left|x+\dfrac{1}{20}\right|+....+\left|x+\dfrac{1}{110}\right|-11x\)
c)Tính gt biểu thức \(C=2x^3-5y^3+2015\) tại x,y thỏa mãn \(\left|x-1\right|+\left(y+2\right)^{20}=0\)
BT9: Tìm x biết
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\)
\(10,\left(x+3\right)^2-x^2=45\)
\(11,\left(5x-4\right)^2-49x^2=0\)
\(12,16\left(x-1\right)^2-25=0\)
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)
\(10,\left(x+3\right)^2-x^2=45\)
\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)
\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)
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