-35 : (2x -1) - 7
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b) Ta có: \(\left(x^2-7\right)\left(x+2\right)-\left(2x-1\right)\left(x-14\right)+x\left(x^2-2x-22\right)+35\)
\(=x^3+2x^2-7x-14-\left(2x^2-28x-x+14\right)+x^3-2x^2-22x+35\)
\(=2x^3-29x+21-2x^2+29x-14\)
\(=2x^3-2x^2+7\)
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35-(x-7)=-7-(-27)+3x
=>35-x+7=-7+27+3x
=>-x+42=20+3x
=>3x-x=42-20
=>2x=22
=>x=11
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a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
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a/ \(\sqrt{x}+\sqrt{x+7}+2\sqrt{x^2+7x}=35-2x\)
\(\Leftrightarrow\sqrt{x}+\sqrt{x+7}+2\sqrt{x\left(x+7\right)}=35-2x\)
Đặt \(a=\sqrt{x}\); \(b=\sqrt{x+7}\) \(\left(a,b\ge0\right)\), ta được:
\(a+b+2ab+2a^2=35\) \(\Leftrightarrow a+2a^2+b+2ab=35\)
\(\Leftrightarrow a\left(1+2a\right)+b\left(1+2a\right)=35\)\(\Leftrightarrow\left(1+2a\right)\left(a+b\right)=35\)
Đến đây bạn chia trường hợp để giải nha
b/ \(P=\frac{1+2x}{1-\sqrt{1+2x}}-\frac{1-2x}{1-\sqrt{1-2x}}\)\(=\frac{\left(1+2x\right)\left(1+\sqrt{1+2x}\right)}{-2x}-\frac{\left(1-2x\right)\left(1+\sqrt{1-2x}\right)}{2x}\)
Tới đây bạn tự làm được k
Câu a ra đến (1+2a)(a+b)=35 rồi giải thế nào vậy bạn. Mình cảm ơn
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\(3x\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)
\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)
\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)
\(=\frac{6}{7}\)
Tìm x
\(a,3x(2x+1)=0\)
\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)
Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)
\(b.\frac{2}{3}-\frac{1}{3}(x-\frac{3}{2})-\frac{1}{2}(2x+1)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-x(\frac{1}{3}+1)=5\)
\(\frac{4}{3}x=\frac{2}{3}-5\)
\(\frac{4}{3}x=\frac{-13}{3}\)
\(x=\frac{-13}{3}\div\frac{4}{3}\)
\(x=\frac{-13}{4}\)
Chúc ban học tốt
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b,x2-2x-35=0
=>(x2-2x+1)-36=0
=>(x-1)2-62=0
=>(x-1-6)(x-1+6)=0
=>(x-7)(x+5)=0
=>x=7 hoặc x=-5
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1:
=>2x-3=0 hoặc 5/2-x=0
=>x=3/2 hoặc x=5/2
2: =>x=1/2+12=12,5
3: =>(2x+3/5-3/5)(2x+3/5+3/5)=0
=>2x(2x+6/5)=0
=>x=0 hoặc x=-3/5
4: =>-1/6x=-1/3
=>x=1/3:1/6=2
5: =>1/4:x=1/4
=>x=1
6: =>2/5x+11/15=1
=>2/5x=4/15
=>x=2/3
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b: =>x-3-5x+15=7
=>-4x+12=7
=>-4x=-5
hay x=5/4
c: =>y(x-7)-2(x-7)=5
=>(x-7)(y-2)=5
\(\Leftrightarrow\left(x-7;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(8;7\right);\left(12;3\right);\left(8;-3\right);\left(2;1\right)\right\}\)
=-35 : ( -2 ) -7
=17 - 7
= 10
à nhầm nó phải như này này
= -35 : (-2) -7
= 17,5-7
=10,5