Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)

\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)

\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)

\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)

\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)

Bài 2:

\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)

\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

Vì \(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)

1/

99-x/101+97-x/103+95-x/105+93-x/107

\(\Leftrightarrow\)(99-x/101+1)+(97-x/103+1)+(95-x/105+1)+(93-x/107+1)=0

\(\Leftrightarrow\)100-x/101+100-x/103+100-x/105+100-x/107=0

\(\Leftrightarrow\)(100-x)(1/101+1/103+1/105+1/107)=0

\(\Leftrightarrow\)100-x=0\(\Leftrightarrow\)

x=-100

Áp dụng tinshh chất dãy tỉ số bằng nhau ; ta được :

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{6+12+25}=\dfrac{86}{43}=2\)

Do đó :

\(\dfrac{x}{3}=2\Rightarrow x=2.3=6\)

\(\dfrac{y}{4}=2\Rightarrow y=2.4=8\)

\(\dfrac{z}{5}=2\Rightarrow z=2.5=10\)

Vậy x = 6 ; y = 8 ; z = 10

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có:

\(\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{6+12+25}=\dfrac{86}{43}=2\) \

\(\Rightarrow x=2.3=6\)

\(y=2.4=8\)

\(z=2.5=10\)

\(\dfrac{x+2}{89}+\dfrac{x+5}{86}>\dfrac{x+8}{83}+\dfrac{x+11}{80}\)

\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}>\dfrac{x+91}{83}+\dfrac{x+91}{80}\)

\(\Leftrightarrow\left(x+91\right)\left(\dfrac{1}{89}+\dfrac{1}{86}\right)>\left(x+91\right)\left(\dfrac{1}{83}+\dfrac{1}{80}\right)\)

Mà \(\dfrac{1}{89}+\dfrac{1}{86}< \dfrac{1}{83}+\dfrac{1}{80}\)

Nên \(x+91< 0\Leftrightarrow x< -91\)

\(\dfrac{x+3}{5}=\dfrac{y-2}{3}=\dfrac{z-1}{7}\) Biết x-y+z = -86

Ta có : \(\dfrac{x+3}{5}=\dfrac{y-2}{3}=\dfrac{z-1}{7}=\dfrac{x+3-\left(y-2\right)+\left(z-1\right)}{5-3+7}=\dfrac{x+3-y+2+z-1}{9}=\dfrac{x-y+z+3+2-1}{9}=\dfrac{-86+4}{9}=\dfrac{-82}{9}\)

+) Với \(\dfrac{x+3}{5}=\dfrac{-82}{9}\)

=> x+3 = \(\dfrac{-82}{9}.5\) = \(\dfrac{-410}{9}\)

=> x = \(\dfrac{-410}{9}\) - 3 = \(\dfrac{-437}{9}\)

+) Với \(\dfrac{y-2}{3}=\dfrac{-82}{9}\)

=> y = \(\dfrac{-76}{3}\) ( Cách làm tương tự phần trên )

+) Với \(\dfrac{z-1}{7}=\dfrac{-82}{9}\)

=> z = \(\dfrac{-565}{9}\) ( Cách làm tương tự phần trên )

Vậy x = \(\dfrac{-437}{9}\)

y = \(\dfrac{-76}{3}\)

z = \(\dfrac{-565}{9}\)

P/s : Đảm bảo kết quả đúng 100% nếu như đề bài không sai

Sửa đề x-y+z = -85 (số này đẹp hơn :))

Áp dụng t/c của dãy tỉ số = nhau có:

\(\dfrac{x+3}{5}=\dfrac{y-2}{3}=\dfrac{z-1}{7}=\dfrac{x+3-y+2+z-1}{5-3+7}=\dfrac{-85+4}{9}=-9\)

\(\Rightarrow\left\{{}\begin{matrix}x+3=-45\\y-2=-27\\z-1=-63\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-48\\y=-25\\z=-62\end{matrix}\right.\)

Vậy...............

P/s: Nếu đề bạn đúng thì thay số vào thôi nhé!

b)\(\dfrac{x+14}{86}+\dfrac{x+15}{85}+\dfrac{x+16}{84}+\dfrac{x+17}{83}+\dfrac{x+116}{4}=0\)

\(\Leftrightarrow\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+1+\dfrac{x+116}{4}-4=0\)

\(\Leftrightarrow\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}+\dfrac{x+100}{4}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x+100=0\).Do \(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\ne0\)

\(\Leftrightarrow x=-100\)

c)\(\dfrac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\dfrac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\dfrac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)

\(\Leftrightarrow\dfrac{1}{\left(x^2+1\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+3\right)}+...+\dfrac{1}{\left(x^2+4\right)\left(x^2+5\right)}=-1\)

\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+2}+\dfrac{1}{x^2+2}-\dfrac{1}{x^2+3}+...+\dfrac{1}{x^2+4}-\dfrac{1}{x^2+5}=-1\)

\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+5}=-1\)\(\Leftrightarrow\dfrac{4}{x^4+6x^2+5}=-1\)

\(\Leftrightarrow\dfrac{x^4+6x^2+9}{x^4+6x^2+5}=0\Leftrightarrow x^4+6x^2+9=0\)

\(\Leftrightarrow\left(x^2+3\right)^2>0\forall x\) (vô nghiệm)

a, x = 99 b, x = -100

c, vo ng

giúp zới

`4/5` tạ `= 4/5 xx 100=80 kg`

`3/5` yến `=3/5 xx 10=6kg`

`4/5` tạ `+3/5` yến `=80+6=86kg`

`->D`

\(\dfrac{4}{5}\) tạ + \(\dfrac{3}{5}\) yến = \(80kg+6kg=86kg\)
\(\rightarrow\)Chọn D

là D

a)vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=>\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)và 2x+3y+5z=86

áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)=\(\dfrac{2x+3y+5z}{6+12+25}\)\(\dfrac{86}{43}\)=2

vì\(\dfrac{2x}{6}\)=2=>2x=2.6=12=>x=12:2=6

\(\dfrac{3y}{12}\)=2=>3y=12.2=24=>y=24:3=8

\(\dfrac{5z}{25}\)=2=>5z=25.2=50=>z=50:5=10

vậy x=6,y=8,z=10

vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)

\(\dfrac{y}{6}\)=\(\dfrac{z}{8}\)=>\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)(2)

từ (1)(2)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)=>\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)và 3x-2y-z=13

áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)=\(\dfrac{3x-2y-z}{27-24-16}\)=\(\dfrac{13}{-13}\)=-1

vì\(\dfrac{3x}{27}\)=-1=>3x=-1.27=-27=>x=-27x;3=-9

\(\dfrac{2y}{24}\)=-1=>2y=-1.24=-24=>y=-24:2=-12

\(\dfrac{z}{16}\)=-1=>z=-1.16=-16

vậy...

\(a,\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}\) và \(2x+3y+5z=86\)

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{6+12+25}=\dfrac{86}{43}=2\)

+) \(\dfrac{2x}{6}=2\Rightarrow2x=2\cdot6=12\Rightarrow x=12:2=6\)

+) \(\dfrac{3y}{12}=2\Rightarrow3y=2\cdot12=24\Rightarrow y=24:3=8\)

+) \(\dfrac{5z}{25}=2\Rightarrow5z=2\cdot25=50\Rightarrow5z=50:5=10\)

Vậy ....

\(b,\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}\left(1\right)\)

\(\dfrac{y}{6}=\dfrac{z}{8}\Leftrightarrow\dfrac{y}{12}=\dfrac{z}{16}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\Leftrightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\) và \(3x-2y-z=13\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)

+) \(\dfrac{3x}{27}=-1\Rightarrow3x=-27\Rightarrow x=-27:3=-9\)

+) \(\dfrac{2y}{24}=-1\Rightarrow2y=-24\Rightarrow y=-24:2=-12\)

+) \(\dfrac{z}{16}=-1\Rightarrow x=-16\)

Vậy .....

\(c,x:y:z=2:5:7\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{3x}{6}=\dfrac{2y}{10}=\dfrac{z}{7}\) và \(3x+2y-z=27\)

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{3x}{6}=\dfrac{2y}{10}=\dfrac{z}{7}=\dfrac{3x+2y-z}{6+10-7}=\dfrac{27}{9}=3\)

+) \(\dfrac{3x}{6}=3\Rightarrow3x=3\cdot6=18\Rightarrow x=18:3=6\)

+) \(\dfrac{2y}{10}=3\Rightarrow2y=3\cdot10=30\Rightarrow y=30:2=15\)

+) \(\dfrac{z}{7}=3\Rightarrow z=3\cdot7=21\)

Vậy ....

d, Xem lại nha!

\(\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+\dfrac{x+16}{4}=4\)

\(\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}=4-\dfrac{x+16}{4}\)

\(\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}\right)=-x\)

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