Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
e) \(\dfrac{x-5}{7}=\dfrac{x+2}{3}\Leftrightarrow3\left(x-5\right)=7\left(x+2\right)\Leftrightarrow3x-15=7x+14\)
\(\Leftrightarrow7x-3x=-15-14\Leftrightarrow4x=-29\Leftrightarrow x=\dfrac{-29}{4}\) vậy \(x=\dfrac{-29}{4}\)
f) \(\left(x-5\right)\left(x+2\right)=\left(x-3\right)\left(x+4\right)\Leftrightarrow x^2+2x-5x-10=x^2+4x-3x-12\)
\(\Leftrightarrow x^2+2x-5x-10-x^2-4x+3x+12=0\Leftrightarrow-4x+2=0\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\)
vậy \(x=\dfrac{1}{2}\)
sao trường khác học nhanh nhỉ? mk còn chưa học căn bậc 2 kìa, toàn lôi máy ra tính ko ak
Mình chỉ làm những câu rõ đề thôi nhé ^^
1/ a/ Đặt \(t=2x-3\) thì pt trở thành \(t^3=\left(t+2\right)^2\Leftrightarrow t^3-t^2-4t-4=0\Leftrightarrow t^2\left(t-1\right)-4\left(t-1\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(t^2-4\right)=0\Leftrightarrow\left(t-2\right)\left(t-1\right)\left(t+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}t=2\\t=1\\t=-2\end{array}\right.\)
Tới đây dễ rồi .
b/ Tương tự đặt \(a=2x-3\) thì pt trở thành \(a^3=a+2\Leftrightarrow a^3-a-2=0\)
Bạn xem lại đề , lớp 7 chưa học giải pt này đâu
c/ VT > 0 => VP > 0 => x > 0
Với x > 0 thì: \(\left|x+3\right|+\left|x+4\right|+\left|x+5\right|=x+3+x+4+x+5=3x+12\)
Tới đây dễ rồi :)
4) |2-|3-2x||=4
<=>\(\left[\begin{array}{nghiempt}2-\left|3-2x\right|=4\\2-\left|3-2x\right|=-4\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}\left|3-2x\right|=-2\left(vl\right)\\\left|3-2x\right|=6\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}3-2x=6\\3-2x=-6\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{9}{2}\end{array}\right.\)
a: \(A\left(x\right)+B\left(x\right)\)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)
\(=8x^2-12x\)
b: C(x)=A(x)-B(x)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)
\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
\(a,\left|2x+3\right|+x=4\)
\(\Rightarrow\left|2x+3\right|=4-x\)
Điều kiện :\(4-x\ge0\Rightarrow x\le4\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=4-x\\2x+3=x-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+x=4-3\\2x-x=-4-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=1\\x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-7\end{matrix}\right.\)
Xét cả 2 trường hợp trên đều thỏa mãn điều kiện
Vậy ...
a) \(\left|\dfrac{1}{2}x\right|=3-2x\)
+) TH1: \(\dfrac{1}{2}x\ge0\Rightarrow x\ge0\)
Khi đó: \(\dfrac{1}{2}x=3-2x\)
\(\Rightarrow x=2\left(3-2x\right)\)
\(\Rightarrow x=6-4x\)
\(\Rightarrow x+4x=6\)
\(\Rightarrow5x=6\Rightarrow x=\dfrac{6}{5}\) (thỏa mãn)
+) TH2: \(-\dfrac{1}{2}x< 0\Rightarrow x>0\)
\(-\dfrac{1}{2}x=3-2x\)
\(\Rightarrow-x=6-4x\)
\(\Rightarrow-x+4x=6\)
\(\Rightarrow3x=6\Rightarrow x=2\) (t/m)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{6}{5}\\x=2\end{matrix}\right.\).
b) \(\left|x-1\right|=3x+2\)
+) TH1: \(x-1\ge0\Rightarrow x\ge1\)
\(x-1=3x+2\)
\(\Rightarrow x-3x=1+2\)
\(\Rightarrow-2x=3\)
\(\Rightarrow x=\dfrac{-3}{2}\) (loại)
+) TH2: \(x-1< 0\Rightarrow x< 1\)
\(-x+1=3x+ 2\)
\(\Rightarrow-x-3x=-1+2\)
\(\Rightarrow-4x=1\)
\(\Rightarrow x=\dfrac{-1}{4}\) (t/m)
Vậy \(x=-\dfrac{1}{4}.\)
c) Tương tự.
a) Vì \(\left|\dfrac{1}{2}x\right|\ge0\)
\(\Leftrightarrow3-2x\ge0\)
Mà \(\left|\dfrac{1}{2}x\right|=3-2x\)
\(\Leftrightarrow\dfrac{1}{2}x=3-2x\)
\(\Leftrightarrow0,5x=3-2x\)
\(\Leftrightarrow0,5x+2x=3\)
\(\Leftrightarrow2,5x=3\)
\(\Leftrightarrow x=\dfrac{3}{2,5}\)
Vậy ................
b) Vì \(\left|x-1\right|\ge0\)
Mà \(\left|x-1\right|=3x+2\)
\(\Leftrightarrow3x+2\ge0\)
\(\Leftrightarrow x-1=3x+2\)
\(\Leftrightarrow x-3x=2+1\)
\(\Leftrightarrow-2x=3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)
Vậy ....................
c) Vì \(\left|5x\right|\ge0\)
Mà \(\left|5x\right|=x+12\)
\(\Leftrightarrow x+12\ge0\)
\(\Leftrightarrow5x=x+12\)
\(\Leftrightarrow5x-x=12\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy ............
Bà tag tui ko có trúng :v