Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)

2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x−1030 +x−1443 +x−595 +x−1488 =0

Trả lời luôn à bạn

a) \(\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=\frac{49}{64}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=\left(\frac{7}{8}\right)^2\\\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=-\left(\frac{7}{8}\right)^2\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{2}{3}:x+\frac{1}{4}=\frac{7}{8}\\\frac{2}{3}:x+\frac{1}{4}=-\frac{7}{8}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{2}{3}:x=\frac{5}{8}\\\frac{2}{3}:x=-\frac{9}{8}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{16}{15}\\x=-\frac{16}{27}\end{array}\right.\)

b) \(\frac{2}{5}:\left|x-2\right|-\frac{1}{3}=\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{5}:\left|x-2\right|=\frac{5}{6}\)

\(\Leftrightarrow\left|x-2\right|=\frac{12}{25}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=\frac{12}{25}\\x-2=-\frac{12}{25}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{62}{25}\\x=\frac{38}{25}\end{array}\right.\)

c) \(\frac{1}{8}-\left(\frac{1}{12}-\frac{1}{3}x\right)^3=0\)

\(\Leftrightarrow\left(\frac{1}{12}-\frac{1}{3}x\right)^3=\frac{1}{8}\)

\(\Leftrightarrow\left(\frac{1}{12}-\frac{1}{3}x\right)^3=\left(\frac{1}{2}\right)^3\)

\(\Leftrightarrow\frac{1}{3}x=-\frac{5}{12}\)

\(\Leftrightarrow x=-\frac{5}{4}\)

Mình đang cần gắp 

bạn còn

a) \(14:\frac{0,4x+0,6}{x}=7\)

\(\frac{0,4x+0,6}{x}=2\)

0,4x + 0,6 = 2.x

2x - 0,4x = 0,6

1,6x = 0,6

x = 0,375

b) \(\left(160\%+\frac{2}{3}x-x\right).12=660\)

\(\left(160\%+\frac{2}{3}x-x\right)=55\)

\(x\left(\frac{2}{3}-1\right)=53,4\)

\(-\frac{1}{3}x=\frac{267}{5}\)

\(x=\frac{267}{5}.\frac{3}{-1}\)

\(x=-160,2\)

c) \(1:\frac{1.2.3.4.....31}{2.2.2.3.2.4.....2.32}=2^x\)

\(1:\frac{1.2.3.4.....31}{2^{31}.2.3.4.....31.2^5}=2^x\)

\(1:\frac{1}{2^{36}}=2^x\)

\(2^{36}=2^x\)

\(x=36\)

chỉ có một số \(\frac{-3}{4}\) thôi nha

ĐKXXD : \(x\ne20;8;3;1\)

\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow\frac{\left(x-1\right)-\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{\left(x-3\right)-\left(x-8\right)}{\left(x-3\right)\left(x-8\right)}+\frac{\left(x-8\right)-\left(x-20\right)}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}+\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow-\frac{1}{x-1}=-\frac{3}{4}\Leftrightarrow x-1=\frac{4}{3}\Rightarrow x=\frac{7}{3}\)

1 

Ez lắm =)

Bài 1:

Với mọi gt \(x,y\in Q\) ta luôn có: 

\(x\le\left|x\right|\) và \(-x\le\left|x\right|\) 

\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)

Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)

Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)

Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)

Dấu "=" xảy ra khi: \(xy\ge0\)

Bài 3: 

Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)

Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a² + b² + c² = 1)

Vậy: (x + y + z)² = x² + y² + z²