Tjm x, biết
\(\left(2:x-\frac{1}{3}\right).\left(x^2-25\right)=0\)
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KC
tjm x\(\in\)tap h0p s0 hưu tj:
\(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
nhanh dj....cam 0n nha
3
MS
6 tháng 6 2016
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
6 tháng 6 2016
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
Y
7 tháng 7 2023
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
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 \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)
5 tháng 9 2016
a)|x+0,573|=2
=>x+0,573=2 hoặc -2
Xét x+0,573=2
=>x=1,427
Xét x+0,573=-2
=>x=-2,573
5 tháng 9 2016
a) | x + 0,573 | = 2
\(\Rightarrow\)x + 0,573 = 2 hoặc x + 0,573 = -2
+) x + 0,573 = 2\(\Rightarrow\)x = 1,427
+) x + 0,573 = -2\(\Rightarrow\)x = -2,573
Vậy x = 1,427 hoặc -2,573
b) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)
\(\Rightarrow x+\frac{1}{3}=3\) hoặc \(x+\frac{1}{3}=-3\)
+) \(x+\frac{1}{3}=3\Rightarrow x=\frac{8}{3}\)
+) \(x+\frac{1}{3}=-3\Rightarrow x=\frac{-10}{3}\)
Vậy \(x=\frac{8}{3}\) hoặc \(x=\frac{-10}{3}\)
Các phần khác làm tương tự nhé bạn
8 tháng 7 2019
1) Tìm x, y biết : \(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)
Ta có :
\(\left|x-y\right|\ge0\forall x;y\)
\(\left|y+\frac{9}{25}\right|\ge0\forall y\)
\(\Rightarrow\left|x-y\right|+\left|y+\frac{9}{25}\right|\ge0\forall x,y\)
\(\Rightarrow\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\Leftrightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+\frac{9}{25}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\frac{9}{25}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-\frac{9}{25}\end{matrix}\right.\)
Vậy : \(x=y=-\frac{9}{25}\)
8 tháng 7 2019
2) Tìm x biết :
a) \(\left|x+\frac{2}{11}\right|>\left|-5,5\right|\)
\(\Rightarrow\left|x+\frac{2}{11}\right|>5,5=\frac{11}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{2}{11}>\frac{11}{2}\\x+\frac{2}{11}>-\frac{11}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\frac{11}{2}-\frac{2}{11}=\frac{117}{22}\\x>-\frac{11}{2}-\frac{2}{11}=-\frac{125}{22}\end{matrix}\right.\Rightarrow x>-\frac{125}{22}\)
Vậy : \(x>-\frac{125}{22}\)
Đúng không ta ? Mình không chắc lắm ....
TT
2 tháng 3 2020
Đặt \(a=24-x,b=x-25\)
Khi đó pt ban đầu trở thành :
\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow30a^2+68ab+30b^2=0\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3a=-5b\\5a=-3b\end{cases}}\)
Đến đây bạn thay vào là dễ rồi nhé ! Chúc bạn học tốt !
7 tháng 7 2019
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
7 tháng 7 2019
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
( 2x - 1/3) . (x2 - 25) = 0
=> 2x - 1/3 = 0 hay x2 - 25 = 0
=> 2x = 0 + 1/3 hay x2 = 0 + 25
=> 2x = 1/3 hay x2 = 25
=. x = 1/6 hay x = 5 hoặc x = -5