Giải phương trình: \((x-5)(x-4)(x-8)(x-10)\)
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Lời giải:
PT \(\Leftrightarrow [(x-5)(x-8)][(x-4)(x-10)]=72x^2\)
\(\Leftrightarrow (x^2-13x+40)(x^2-14x+40)=72x^2\)
Đặt \(x^2-13x+40=a\) thì pt trở thành:
\(a(a-x)=72x^2\)
\(\Leftrightarrow a^2-ax-72x^2=0\)
\(\Leftrightarrow a^2-9ax+8ax-72x^2=0\)
\(\Leftrightarrow a(a-9x)+8x(a-9x)=0\)
\(\Leftrightarrow (a-9x)(a+8x)=0\)
Nếu $a-9x=0$
\(\Leftrightarrow x^2-13x+40-9x=0\)
\(\Leftrightarrow x^2-22x+40=0\)
\(\Leftrightarrow (x-2)(x-20)=0\Rightarrow \left[\begin{matrix} x=2\\ x=20\end{matrix}\right.\)
Nếu $a+8x=0$
\(\Leftrightarrow x^2-13x+40+8x=0\)
\(\Leftrightarrow x^2-5x+40=0\Leftrightarrow (x-\frac{5}{2})^2=-\frac{135}{4}\) (vô lý)
Vậy........
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|x-9|=2x+5
Xét 3 TH
TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)
TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)
TH3: x=9 =>0=23(L)
Vậy x= 4/3
Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)
\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)
\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)
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a: Ta có: \(\sqrt{x^2-x+3}+7=10\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)
\(\Leftrightarrow x^2-4x+8=4\)
\(\Leftrightarrow x-2=0\)
hay x=2
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(x+1)(x+2)(x+4)(x+8)=28x2
\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+9x+8\right)=28x^2\)(1)
Thấy x=0 không là nghiệm của (1). CHia 2 vế (1) cho x2 ta đc:
\(\left(1\right)\Leftrightarrow\left(x+\frac{8}{x}+6\right)\left(x+\frac{8}{9}+9\right)=28\)
Đặt \(t=x+\frac{8}{x}\)ta có:
\(\left(1\right)\Rightarrow\left(t+6\right)\left(t+9\right)=28\)
\(\Leftrightarrow t^2+15t+26=0\Leftrightarrow\orbr{\begin{cases}t=-2\\t=-13\end{cases}}\)
- Với \(t=-2\Rightarrow x+\frac{8}{x}=-2\Leftrightarrow x^2+2x+8=0\Leftrightarrow\left(x+1\right)^2+7>0\)(vô nghiệm)
- Với \(t=-13\Rightarrow x+\frac{8}{x}=-13\Rightarrow x^2+13x+8=0\)
\(\Delta=13^2-4\left(1.8\right)=137\)\(\Rightarrow x_{1,2}=\frac{-13\pm\sqrt{137}}{2}\)(thỏa mãn)
Vậy...
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câu a bài 1:(2x+1)(3x-2)=(5x-8)(2x+1)
<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0
<=>(2x+1)(3x-2-5x+8)=0
<=>(2x+1)(6-2x)=0
bước sau tự làm nốt nha !
câu b:gợi ý: tách 4x^2-1thành (2x-1)(2x+1) rồi làm như câu a
Đặng Thị Vân Anh tuy mk k cần nx nhưng dù s cx cảm ơn bn nha :)
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\(\dfrac{x+2}{89}+\dfrac{x+5}{86}>\dfrac{x+8}{83}+\dfrac{x+11}{80}\)
\(\Leftrightarrow\dfrac{x+2}{89}+1+\dfrac{x+5}{86}+1>\dfrac{x+8}{83}+1+\dfrac{x+11}{80}+1\)
\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}>\dfrac{x+91}{83}+\dfrac{x+91}{80}\)
\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}-\dfrac{x+91}{83}-\dfrac{x+91}{80}>0\)
\(\Leftrightarrow\left(x+91\right)\left(\dfrac{1}{89}+\dfrac{1}{86}-\dfrac{1}{83}-\dfrac{1}{80}\right)>0\)
Ta có: \(\dfrac{1}{89}+\dfrac{1}{86}+\dfrac{1}{83}+\dfrac{1}{80}< 0\)
\(\Leftrightarrow x+91< 0\)
\(\Leftrightarrow x< -91\)
Vậy...........
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\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{8}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+3\left(\dfrac{1}{8}-\dfrac{1}{y}\right)=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+\dfrac{3}{8}-\dfrac{3}{x}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{1}{x}=\dfrac{1}{24}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\x=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{12}\\x=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=24\end{matrix}\right.\)
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ĐKXĐ: ...
\(\Leftrightarrow\frac{9\left(2x+5\right)^2}{4\left(x+4\right)^2}+\left(2x+5\right)^2=8\)
\(\Leftrightarrow\frac{9\left(2x+5\right)^2}{4\left(x+4\right)^2}-2.\frac{3\left(2x+5\right)}{2\left(x+4\right)}.\left(2x+5\right)+\left(2x+5\right)^2+\frac{3\left(2x+5\right)^2}{x+4}=8\)
\(\Leftrightarrow\left(\left(2x+5\right)-\frac{3\left(2x+5\right)}{2\left(x+4\right)}\right)^2+\frac{3\left(2x+5\right)^2}{x+4}=8\)
\(\Leftrightarrow\left(\frac{\left(2x+5\right)^2}{2\left(x+4\right)}\right)^2+\frac{3\left(2x+5\right)^2}{x+4}-8=0\)
Đặt \(\frac{\left(2x+5\right)^2}{x+4}=a\)
\(\Leftrightarrow\frac{a^2}{4}+3a-8=0\)
Nghiệm xấu, bạn tự giải nốt