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1: Trường hợp 1: x<-2
Pt sẽ là -x-2+5-x=7
=>-2x+3=7
=>-2x=4
hay x=-2(loại)
Trường hợp 2: -2<=x<5
Pt sẽlà x+2+5-x=7
=>7=7(luôn đúng)
Trường hợp 3: x>=5
Pt sẽ là x+2+x-5=7
=>2x-3=7
=>x=5(nhận)
4: \(\left|x^2-2x\right|=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(x^2-2x\right)^2=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x^2-2x-x\right)\left(x^2-2x+x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x^2-3x\right)\left(x^2-x\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;1;3\right\}\)
5: Ta có: \(\left|2x+3\right|=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(2x+3+x+2\right)\left(2x+3-x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x+5\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{3};-1\right\}\)
6: |5x-4|=|x+2|
=>5x-4=x+2 hoặc 5x-4=-x-2
=>4x=6 hoặc 6x=2
=>x=3/2 hoặc x=1/3
a, (ko vt lại đề)
=> -5x- 1-1/2x -1/3=3/2x -5/6
=> -5x - 1/2x +3/2x = 1+1/3 - 5/6
=>( -5 -1/2 + 3/2 )x =1/2
=> -4x = 1/2
=> x = -1/8
Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)
\(x+\frac{4}{5}=\pm\frac{4}{5}\)
\(TH1:x+\frac{4}{5}=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)
\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)
\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)
Vậy x ∈ {0; \(\frac{-8}{5}\)}
a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)
1) \(\frac{17}{6}-\left(x-\frac{7}{6}\right)=\frac{7}{4}\)
\(\Rightarrow x-\frac{7}{6}=\frac{17}{6}-\frac{7}{4}\)
\(\Rightarrow x=\frac{13}{12}+\frac{7}{6}=\frac{9}{4}\)
2) \(\frac{3}{35}-\left(\frac{3}{5}-x\right)=\frac{2}{7}\)
\(\Rightarrow\)\(\frac{3}{5}-x=\frac{3}{35}-\frac{2}{7}=-\frac{1}{5}\)
\(\Rightarrow x=\frac{3}{5}-\left(-\frac{1}{5}\right)=\frac{4}{5}\)
3) 4) Hjhj^_^^_^
Theo đề bài ta có \(M(x) = 2{x^4} - 5{x^3} + 7{x^2} + 3x\)
\(\begin{array}{l}M(x) + Q(x) = 6{x^5} - {x^4} + 3{x^2} - 2\\ \Rightarrow Q(x) = (6{x^5} - {x^4} + 3{x^2} - 2) - (2{x^4} - 5{x^3} + 7{x^2} + 3x)\\ \Rightarrow Q(x) = 6{x^5} - {x^4} + 3{x^2} - 2 - 2{x^4} + 5{x^3} - 7{x^2} - 3x\\Q(x) = 6{x^5} - 3{x^4} + 5{x^3} - 4{x^2} - 3x - 2\end{array}\)
Theo đề bài ta có :
\(\begin{array}{l}N(x) - M(x) = - 4{x^4} - 2{x^3} + 6{x^2} + 7\\ \Rightarrow N(x) = - 4{x^4} - 2{x^3} + 6{x^2} + 7 + 2{x^4} - 5{x^3} + 7{x^2} + 3x\\ \Rightarrow N(x) = - 2{x^4} - 7{x^3} + 13{x^2} + 3x + 7\end{array}\)