giai cac phuong trinh sau \(\left|x+2\right|+\left|x+9\right|+\left|x+2011\right|=4x\)
Những câu hỏi liên quan
giai phuong trinh \(\left|x+2\right|^{2010}+\left|x+3\right|^{2011}=1\)
Nhận thấy \(\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\) là nghiệm của pt
- Với \(x>-2\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|>0\\\left|x+3\right|>1\end{matrix}\right.\) \(\Rightarrow\left|x+2\right|^{2010}+\left|x+3\right|^{2011}>1\)
\(\Rightarrow\) pt vô nghiệm
- Với \(x< -3\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|>1\\\left|x+3\right|>0\end{matrix}\right.\) \(\Rightarrow\left|x+2\right|^{2010}+\left|x+3\right|^{2011}>1\)
\(\Rightarrow\) pt vô nghiệm
- Với \(-3< x< -2\Rightarrow\left\{{}\begin{matrix}\left|x+3\right|< 1\\\left|x+2\right|< 1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|^{2010}< \left|x+2\right|\\\left|x+3\right|^{2011}< \left|x+3\right|\end{matrix}\right.\) \(\Rightarrow VT< \left|x+2\right|+\left|x+3\right|=-x-2+x+3=1\)
\(\Rightarrow\) pt vô nghiệm
Vậy pt có đúng 2 nghiệm \(\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Giai phuong trinh sau :
c ) \(\left(2-x\right)\left(2x-1\right)+\left(4x^2-4x+1\right)=0\)
\(\left(2-x\right)\left(2x-1\right)+\left(4x^2-4x+1\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(2x-1\right)+\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2-x+2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=0\\x+1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-1\end{array}\right.\)
Vậy phương trình có tập nghiệm \(\left\{-1;\frac{1}{2}\right\}\)
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(2-x)(2x-1)+(4x^2-4x+1)=0
Ta có: (2x-1)(2-x)+(2x-1)^2=0
(2x-1)(2-x+2x-1)=0
Sau đó bn tự lam nha tại vì mk làm bằng phone
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giai he phuong trinh \(\left\{\left|x-2\right|+2\left|y-1\right|=9\right\}\left\{x+\left|y-1\right|=-1\right\}\)
giai cac phuong trinh sau:
a.\(\frac{6}{\left(x+1\right)\left(x+2\right)}+\frac{8}{\left(x-1\right)\left(x+4\right)}=1\)
b.\(x^3+\frac{1}{x^3}=13\left(x+\frac{1}{x}\right)\)
giai phuong trinh \(\left(x^2-4x+3\right)\left(x^2-6x+8\right)=8\)
\(\left(x^2-4x+3\right)\left(x^2-6x+8\right)=8\)
\(\left(x^2-3x-x+3\right)\left(x^2-4x-2x+8\right)=8\)
\(\left[x\left(x-3\right)-1\left(x-3\right)\right]\left[x\left(x-4\right)-2\left(x-4\right)\right]=8\)
\(\left(x-1\right)\left(x-3\right)\left(x-2\right)\left(x-4\right)=8\)
\(\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=8\)
\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-8=0\)
Đặt \(t=x^2-5x+4\)
\(t\left(t+2\right)-8=0\)
\(t^2+2t-8=0\)
\(t^2+4t-2t-8=0\)
\(t\left(t+4\right)-2\left(t+4\right)=0\)
\(\left(t+4\right)\left(t-2\right)=0\)
\(\orbr{\begin{cases}t+4=0\\t-2=0\end{cases}}\)
\(\orbr{\begin{cases}t=-4\\t=2\end{cases}}\)
\(\orbr{\begin{cases}x^2-5x+4=-4\\x^2-5x+4=2\end{cases}}\)
\(\orbr{\begin{cases}x^2-5x+8=0\left(ptvn\right)\\x^2-5x+2=0\end{cases}}\)
\(x^2-5x+2=0\)
\(\orbr{\begin{cases}x=\frac{5+\sqrt{17}}{2}\\x=\frac{5-\sqrt{17}}{2}\end{cases}}\)
\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)0
Giai phuong trinh
\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)
<=> \(3\left(x^2+1\right)^2.4+3\left(x^2+x\right).4-9.4=0.4\)
<=> \(3\left(x^2+1\right)^2+12\left(x^2+x\right)-36=0\)
<=> \(3x^4+18x^2+12x-33=0\)
<=> \(3\left(x-1\right)\left(x^3+x^2+7x+11\right)=0\)
<=> \(x-1=0\)
<=> \(x=1\)
Mà vì: \(x^3+x^2+7x+11\ne0\)
=> x = 1
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\(=>\frac{3}{4}\left[\left(x^2+1\right)^2+4\left(x^2+1\right)+4\right]-12=0\)
\(=>\frac{3}{4}\left(x^2+1+2\right)^2-12=0\)
\(=>\left(x^2+3\right)^2=16\)
Đến đây tự tìm nha
Hok tốt
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Giai he phuong trinh:
a) \(\hept{\begin{cases}\left(x+y\right).\left(y+z\right)=187\\\left(y+z\right).\left(z+x\right)=154\\\left(z+x\right).\left(x+y\right)=238\end{cases}}\)
b) \(\hept{\begin{cases}x^2-y^2=1\\4x^2-5xy=2\end{cases}}\)
\(Taco:\)
\(\left(x+y\right)\left(y+z\right)=187\Leftrightarrow xy+xz+yy+yz=187\)
\(\left(y+z\right)\left(z+x\right)=154\Leftrightarrow yz+xy+zz+xz=154\)
\(\left(z+x\right)\left(x+y\right)=238\Leftrightarrow xz+zy+xx+xy=238\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)+\left(x+z\right)\left(x+y\right)+\left(y+z\right)\left(z+x\right)=579\)
\(\Leftrightarrow xy+zx+yy+yz+yz+xy+zz+xz+xz+zy+xx+xy=579\)
\(\Leftrightarrow3\left(xz+xy+yz\right)+x^2+y^2+z^2=579\)
\(\left(z+x\right)\left(x+y\right)-\left(x+y\right)\left(y+z\right)=51\)
\(\Leftrightarrow\left(x+y\right)\left(x-y\right)=x^2-y^2=51\)
\(\left(z+x\right)\left(x+y\right)-\left(y+z\right)\left(x+z\right)=84\)
\(\Leftrightarrow\left(x+z\right)\left(x-z\right)=84\Leftrightarrow x^2-z^2=84\)
\(\Leftrightarrow y^2-z^2=33\)
đến đây tịt
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giai phuong trinh\(\left(x+5\right)+\left(x-5\right)+\left(x.5\right)+\left(x\div5\right)=180\)
\(\left(x+5\right)+\left(x-5\right)+5x+x\div5=180\)
\(\Leftrightarrow\left(x+x+5x\right)+\left(5-5\right)+\frac{x}{5}=180\)
\(\Leftrightarrow7x+0+\frac{x}{5}=180\)
\(\Leftrightarrow7x+\frac{x}{5}=180\)
\(\Leftrightarrow\frac{35x+x}{5}=180\)
\(\Leftrightarrow35x+x=180.5\)
\(\Leftrightarrow36x=900\)
\(\Leftrightarrow x=\frac{900}{36}\)
\(\Leftrightarrow x=25\)
Vậy phương trình có 1 nghiệm duy nhất là 25
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(x + 5) + (x - 5) + 5x + \(\frac{x}{5}\)= 180
<=> x + 5 + x - 5 + 5x + \(\frac{x}{5}\) = 180
<=> 7x + \(\frac{x}{5}\) = 180
<=> \(\frac{36x}{5}=180\)
\(\Leftrightarrow x=\frac{180.5}{36}=25\)
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\(\left(x+5\right)+\left(x-5\right)+\left(x.5\right)+\left(x:5\right)=180\)\(\Leftrightarrow2x+5x+\frac{x}{5}=180\Leftrightarrow7x+\frac{x}{5}=180\)
\(\Leftrightarrow\frac{35x+x}{5}=\frac{900}{5}\Leftrightarrow35x+x=900\Leftrightarrow36x=900\Leftrightarrow x=25\)
Vậy phương trình có tập nghiệm S = { 25 }
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giai cac he phuong trinh sau
15) \(\left\{{}\begin{matrix}3x+2y=7\\x^2+y^2-7x+xy=0\end{matrix}\right.\)
16)\(\left\{{}\begin{matrix}2x+3y=5\\x^2+xy+y^2-4x=-1\end{matrix}\right.\)
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