Tìm x :
a. \(\left|5-\left|x-4\right|+2x\right|=10\)
b. -\(\left|7x-10\right|=2x+1\)
Những câu hỏi liên quan
1. dfrac{5left(x-1right)+2}{6}-dfrac{7x-1}{4}dfrac{2left(2x+1right)}{7}-5
2. x-dfrac{3left(x+30right)}{15}-24dfrac{1}{2}dfrac{7x}{10}-dfrac{2left(10x+2right)}{5}
3. 14dfrac{1}{2}-dfrac{2left(x+3right)}{5}dfrac{3x}{2}-dfrac{2left(x-7right)}{3}
4. dfrac{x+1}{3}+dfrac{3left(2x+1right)}{4}dfrac{2x+3left(x+1right)}{6}+dfrac{7+12x}{12}
5. dfrac{3left(2x-1right)}{4}-dfrac{3x+1}{10}+1dfrac{2left(3x+2right)}{5}
6. x-dfrac{3}{17}left(2x-1right)dfrac{7}{34}left(1-2xright)+dfrac{10x-3}{2}
7. dfrac{3le...
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1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
rut gọn biểu thức sau
a)\(3x^2-2x\left(5+1.5x\right)+10\)
b)\(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
c)\(\hept{\begin{cases}\\\end{cases}}2x-3\left(x-1\right)-5\orbr{\begin{cases}\\\end{cases}x-4\left(3-2x\right)+10\orbr{\begin{cases}\\\end{cases}}}\)
Giải phương trình:
1. frac{2x+3}{4}-frac{5x+3}{6}frac{3-4x}{12}
2. frac{3.left(2x+1right)}{4}-1frac{15x-1}{10}
3. frac{2x-1}{5}-frac{x-2}{3}frac{x+7}{15}
4. frac{x+3}{2}-frac{x-1}{3}frac{x+5}{6}+1
5. frac{x-4}{5}-frac{3x-2}{10}-xfrac{2x-5}{3}-frac{7x+2}{6}
6. frac{left(x+2right)left(x+10right)}{3}-frac{left(x+4right)left(x+10right)}{12}frac{left(x-2right)left(x+4right)}{4}
7. frac{left(x+2right)^2}{8}-2left(2x-1right)25+frac{left(x-2right)^2}{8}
8.frac{7x^2-14x-5}{5}frac{left(2x+1right)^...
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Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
Chứng minh biểu thức sau không phụ thuộc vào giá trị của biến :
\(A=x.\left(5x-3\right)-x^2.\left(x-1\right)+x.\left(x^2-6x\right)-10+3x+x.\left(x^2+x+1\right)-x^2.\left(x+1\right)-x+5\)
\(B=3.\left(2x-1\right)-5.\left(x-3\right)+6.\left(3x-4\right)-19x+x.\left(3x+12\right)-\left(7x-20\right)+x^2.\left(2x-3\right)-x.\left(2x^2+5\right)\)
T ko biết làm, chỉ hỏi liên thiên thôi :)))
Hủ phải không???? OvO Dưa Trong Cúc
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a: \(A=5x^2-3x-x^3+x^2+x^3-6x^2-10x+3x+x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)
\(=-10x+x^3+x^2+x-x^3-x^2-x+5\)
=-11x+5
b: \(=6x-3-5x+15+18x-24-19x+3x^2+12x-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
\(=3x^2+12x-12-7x+20+2x^3-3x^2-2x^3-5x\)
\(=8\)
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Tìm x a,3-xx+1,8b,2x-57x+35c,2left(x+10right)3left(x-6right)d,8left(x-dfrac{3}{8}right)+16left(dfrac{1}{6}+xright)+xe,dfrac{2}{9}-3xdfrac{4}{3}-xg,dfrac{1}{2}x+dfrac{5}{6}dfrac{3}{4}x-dfrac{1}{2}h,x-4dfrac{5}{6}left(6-dfrac{6}{5}xright)k,7x^2-116x^2-2m,5left(x+3.2^3right)10^2n,dfrac{4}{9}-(dfrac{1}{6^2})dfrac{2}{3}left(x-dfrac{2}{3}right)^2+dfrac{5}{12}
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Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
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\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
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a: 3-x=x+1,8
=>-2x=-1,2
=>x=0,6
b: 2x-5=7x+35
=>-5x=40
=>x=-8
c: 2(x+10)=3(x-6)
=>3x-18=2x+20
=>x=38
d; 8(x-3/8)+1=6(1/6+x)+x
=>8x-3+1=1+6x+x
=>8x-2=7x+1
=>x=3
e: =>-3x+x=4/3-2/9
=>-2x=12/9-2/9=10/9
=>x=-5/9
g: =>3/4x-1/2x=5/6+1/2
=>1/4x=5/6+3/6=8/6=4/3
=>x=4/3*4=16/3
h: =>x-4=-x+5
=>2x=9
=>x=9/2
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1. \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
2 . \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
3 . \(4\left(3x-2\right)-3\left(x-4\right)=7x+10\)
4. \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}
TÌM MAX; MIN 1. -x^2-y^2+xy+2x+2y2. left(x-2right)left(x-5right)left(x^2-7x-10right)3.left|x-4right|left(2-left|x-4right|right)4. left(2x-1right)^2-3left|2x-1right|+25. left(x-1right)left(x+2right)left(x+3right)left(x+6right)6. left(x+1right)left(x+2right)left(x+3right)left(x+4right)
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TÌM MAX; MIN
1. \(-x^2-y^2+xy+2x+2y\)
2. \(\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)\)
3.\(\left|x-4\right|\left(2-\left|x-4\right|\right)\)
4. \(\left(2x-1\right)^2-3\left|2x-1\right|+2\)
5. \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
6. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
Giải các phương trình sau :
a) dfrac{5left(x-1right)+2}{6}-dfrac{7x-1}{4}dfrac{2left(2x+1right)}{7}-5
b) dfrac{3left(x-3right)}{4}+dfrac{4x-10,5}{10}dfrac{3left(x+1right)}{5}+6
c) dfrac{2left(3x+1right)+1}{4}-5dfrac{2left(3x-1right)}{5}-dfrac{3x+2}{10}
d) dfrac{x+1}{3}+dfrac{3left(2x+1right)}{4}dfrac{2x+3left(x+1right)}{6}+dfrac{7+12x}{12}
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Giải các phương trình sau :
a) \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
b) \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
c) \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
d) \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
giải phương trình a.\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
b.\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)
Câu 1:
ĐK: \(x\geq -2\)
Đặt \(\sqrt{x+5}=a; \sqrt{x+2}=b(a,b\geq 0)\)
\(\Rightarrow ab=\sqrt{(x+5)(x+2)}=\sqrt{x^2+7x+10}\)
PT trở thành:
\((a-b)(1+ab)=3\)
\(\Leftrightarrow (a-b)(1+ab)=(x+5)-(x+2)=a^2-b^2\)
\(\Leftrightarrow (a-b)(1+ab)-(a-b)(a+b)=0\)
\(\Leftrightarrow (a-b)(1+ab-a-b)=0\)
\(\Leftrightarrow (a-b)(a-1)(b-1)=0\)
Vì \(a\neq b\Rightarrow \left[\begin{matrix} a-1=0\\ b-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} a=\sqrt{x+5}=1\\ b=\sqrt{x+2}=1\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-4\\ x=-1\end{matrix}\right.\). Vì $x\geq -2$ nên chỉ có $x=-1$ là nghiệm duy nhất.
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Câu 2:
ĐK: \(-4\leq x\leq 4\)
Ta có: \((\sqrt{x+4}-2)(\sqrt{4-x}+2)=2x\)
\(\Leftrightarrow \frac{(x+4)-2^2}{\sqrt{x+4}+2}.(\sqrt{4-x}+2)=2x\)
\(\Leftrightarrow x.\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}=2x\)
\(\Leftrightarrow x\left(\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}-2\right)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ \sqrt{4-x}+2=2\sqrt{x+4}+4(*)\end{matrix}\right.\)
Xét $(*)$
Đặt \(\sqrt{4-x}=a; \sqrt{x+4}=b\) thì ta có hệ:
\(\left\{\begin{matrix} a^2+b^2=8\\ a+2=2b+4\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a^2+b^2=8\\ a=2(b+1)\end{matrix}\right.\)
\(\Rightarrow 4(b+1)^2+b^2=8\)
\(\Leftrightarrow 5b^2+8b-4=0\Leftrightarrow (5b-2)(b+2)=0\)
\(\Rightarrow b=\frac{2}{5}\) (do \(b\geq 0)\)
\(\Rightarrow x+4=b^2=\frac{4}{25}\Rightarrow x=\frac{-96}{25}\) (t/m)
Vậy \(x\in \left\{ \frac{-96}{25}; 0\right\}\)
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