Tìm x biết
a) x:8=5:4
b) 0,6:x=x:2,4
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Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(a,\text{Với }x< -2\Rightarrow3-x-x-2=4\\ \Rightarrow-2x=3\Rightarrow x=-\dfrac{3}{2}\left(ktm\right)\\ \text{Với }-2\le x< 3\Rightarrow3-x+x+2=4\\ \Rightarrow0x=-1\Rightarrow x\in\varnothing\\ \text{Với }x\ge3\Rightarrow x-3+x+2=4\\ \Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\left(ktm\right)\)
Vậy \(x\in\varnothing\)
\(b,\text{Với }x< 2\Rightarrow4-2x+18-6x=21\\ \Rightarrow22-8x=21\Rightarrow x=\dfrac{1}{8}\left(tm\right)\\ \text{Với }2\le x< 3\Rightarrow2x-4+18-6x=21\\ \Rightarrow-4x+14=21\Rightarrow x=-\dfrac{7}{4}\left(ktm\right)\\ \text{Với }x\ge3\Rightarrow2x-4+6x-18=21\\ \Rightarrow8x=43\Rightarrow x=\dfrac{43}{8}\left(tm\right)\)
Vậy \(x\in\left\{\dfrac{1}{8};\dfrac{43}{8}\right\}\)
a: \(\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)
\(a,3\cdot x-15=x+35\)
\(\Rightarrow3x-x=35+15\)
\(\Rightarrow 2x=50\)
\(\Rightarrow x = 50:2\)
\(\Rightarrow x= 25\)
\(b,(8x-16)(x-5)=0\)
\(+, TH1: 8x-16=0\)
\(\Rightarrow8x=16\)
\(\Rightarrow x = 16:8\)
\(\Rightarrow x=2\)
\(+,TH2: x-5=0\)
\(\Rightarrow x =5\)
\(c,x(x+1)=2+4+6+8+10+...+2500\) \(^{\left(1\right)}\)
Đặt \(A=2+4+6+8+10+...+2500\)
Số các số hạng của \(A\) là: \(\left(2500-2\right):2+1=1250\left(số\right)\)
Tổng \(A\) bằng: \(\left(2500+2\right)\cdot1250:2=1563750\)
Thay \(A=1563750\) vào \(^{\left(1\right)}\), ta được:
\(x\left(x+1\right)=1563750\)
\(\Rightarrow x\left(x+1\right)=1250\cdot1251\)
\(\Rightarrow x =1250\)
#\(Toru\)
\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)
\(a.\)
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-2\right)\cdot2=-4\\y=\left(-2\right)\cdot5=-10\end{matrix}\right.\)
\(b.\)
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot7=28\\y=5\cdot4=20\end{matrix}\right.\)
\(a,\dfrac{5}{8}=\dfrac{x}{14}\)
\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)
Vậy \(x=8,75\)
\(b,\dfrac{x}{6}=-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{1.6}{3}=-2\)
Vậy \(x=-2\)
\(c,-\dfrac{3}{5}=\dfrac{x}{10}\)
\(\Rightarrow x=-\dfrac{3.10}{5}=-6\)
Vậy \(x=-6\)
câu d đã có đáp án
Tìm x biết:
a) x:8=5:4
x:8=1,25
x =1,25x8
x =10
b) 0,6 :x=x:2,4
x2=0,6x2,4
x2=1,44\(x=\sqrt{1,44}=1,2\)