Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Rightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\left(x-7\right)^{x+1}\left[1^2-\left(x-7\right)^{5^2}\right]=0\)
\(\Rightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^5\right]\left[1+\left(x-7\right)^5\right]=0\)
=>(x-7)x+1=0 hoặc 1-(x-7)5=0 hoặc 1+(x-7)5=0
+)Nếu (x-7)x+1=0
=>x-7=0
=>x=7
+)Nếu 1-(x-7)5=0
=>(x-7)5=1
=>x-7=1
=>x=8
+)Nếu 1+(x-7)5=0
=>(x-7)5=-1
Vì \(\left(x-7\right)^5\ge0\) với mọi x
=>không tìm được x thỏa mãn 1+(x-7)5=0
Vậy x=7 hoặc x=8
\(a)\) \(\left(2x+1\right)\left(2x-3\right)=7\)
Có \(4\) trường hợp :
\(\Leftrightarrow\)\(\hept{\begin{cases}2x+1=1\\2x-3=7\end{cases}\Leftrightarrow\hept{\begin{cases}2x=0\\2x=10\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x=5\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}2x+1=-1\\2x-3=-7\end{cases}\Leftrightarrow\hept{\begin{cases}2x=-2\\2x=-4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\x=-2\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}2x+1=7\\2x-3=1\end{cases}\Leftrightarrow\hept{\begin{cases}2x=6\\2x=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\x=2\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}2x+1=-7\\2x-3=-1\end{cases}\Leftrightarrow\hept{\begin{cases}2x=-8\\2x=2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-4\\x=1\end{cases}}}\)
Vậy không có giá trị nào của x thoả mãn đề bài
\(b)\) \(x\left(x-7\right)+3\left(x-7\right)=11\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x-7\right)=11\)
Có \(4\) trường hợp :
\(\Leftrightarrow\)\(\hept{\begin{cases}x+3=1\\x-7=11\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\x=18\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x+3=-1\\x-7=-11\end{cases}\Leftrightarrow\hept{\begin{cases}x=-4\\x=-4\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x+3=11\\x-7=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=8\\x=8\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x+3=-11\\x-7=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=-14\\x=6\end{cases}}}\)
Vậy \(x\in\left\{-4;8\right\}\)
\(a.\left(x-4\right)\left(x+7\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-4=0\\x+7=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=-7\end{cases}}}\)
\(b.x\left(x+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=-3\end{cases}}}\)
\(c.\left(x-2\right)\left(5-x\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-2=0\\5-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=5\end{cases}}}\)
\(d.\left(x-1\right)\left(x^2+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x^2=-1\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\x=-\left(-1\right)or\left(-1\right)\end{cases}}}\)
a) ( x - 4 ) . ( x + 7 ) = 0
một phép nhân có tích bằng 0
=> một trong hai thừa số này bằng 0
+) nếu x - 4 = 0 => x = 0 + 4 = 4
+) nếu x + 7 = 0 => x = 0 - 7 = -7
vậy x = { 4 ; -7 }
b) x . ( x + 3 ) = 0
x + 3 = 0 : x
x + 3 = 0
x = 0 - 3
x = -3
vậy x = -3
c) ( x - 2 ) . ( 5 - x ) = 0
một phép nhân có tích bằng 0
=> một trong hai thừa số này bằng 0
+) nếu x - 2 = 0 => x = 0 + 2 = 2
+) nếu 5 - x = 0 => x = 5 - 0 = 5
vậy x = { 2 ; 5 }
d) ( x - 1 ) . ( x2 + 1 ) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
+) x - 1 = 0 => x = 0 + 1 = 1
+) x2 + 1 = 0 => x2 = 0 - 1 = -1 => x = -1
vậy x = { 1 ; -1 }
a) \(\left(x-4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=7\end{array}\right.\)
b) \(x\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\end{array}\right.\)
c) \(\left(x-2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\5-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=5\end{array}\right.\)
d) \(\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x-1=0\) ( Vì \(x^2+1>0\) )
\(\Leftrightarrow x=1\)
a)
\(\left(x-4\right)\left(x-7\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=7\end{array}\right.\)
Vậy x = 4 ; x = 7
b)
\(x\left(x+3\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\end{array}\right.\)
Vậy x = 0 ; x = - 3
c)
\(\left(x-2\right)\left(5-x\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=5\end{array}\right.\)
Vậy x = 2 ; x = 5
d)
\(\left(x-1\right)\left(x^2+1\right)=0\)
Mà \(x^2+1\ge1\)
=> x = - 1
Vậy x = - 1
a) (x + 1/2) . (2/3 − 2x) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)
\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)
\(\Rightarrow x=-\frac{2}{11}\)
c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)
\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)
\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)
\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)
\(\Rightarrow x=1\)
d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)
\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)
\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)
\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)
\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)
\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)
\(\Rightarrow x\approx28,7\) (số hơi lẻ)
e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
bài toán này cực khó
Bài này lớp 6 ak khó nhỉ