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Bài 9:
a: ĐKXĐ: \(\begin{cases}5-2x\ge0\\ x-1\ge0\end{cases}\Rightarrow1\le x\le\frac52\)
\(\sqrt{5-2x}=\sqrt{x-1}\)
=>5-2x=x-1
=>-2x-x=-1-5
=>-3x=-6
=>x=2(nhận)
b: ĐKXĐ: \(\begin{cases}x^2-3x+2\ge0\\ x-1\ge0\end{cases}\Rightarrow\begin{cases}\left(x-2\right)\left(x-1\right)>=0\\ x-1\ge0\end{cases}\)
=>(x>=2 hoặc x<=1) hoặc x>=1
=>(x>=2 hoặc x=1)
\(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
=>\(x^2-3x+2=x-1\)
=>(x-1)(x-2)-(x-1)=0
=>(x-1)(x-3)=0
=>\(\left[\begin{array}{l}x-1=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=3\left(nhận\right)\end{array}\right.\)
c: ĐKXĐ: \(\begin{cases}x^2-3x+2\ge0\\ 5x+2\ge0\end{cases}\)
=>\(\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\ 5x+2\ge0\end{cases}\)
=>(x>=2 hoặc x<=1) và x>=-2/5
=>x>=2 hoặc -2/5<=x<=1
\(\sqrt{x^2-3x+2}-\sqrt{5x+2}=0\)
=>\(\sqrt{x^2-3x+2}=\sqrt{5x+2}\)
=>\(x^2-3x+2=5x+2\)
=>\(x^2-8x=0\)
=>x(x-8)=0
=>x=0(nhận) hoặc x=8(nhận)
d: ĐKXĐ: \(\begin{cases}3x+7\ge0\\ x+1\ge0\end{cases}=>x\ge-1\)
\(\sqrt{3x+7}-\sqrt{x+1}=0\)
=>\(\sqrt{3x+7}=\sqrt{x+1}\)
=>3x+7=x+1
=>2x=-6
=>x=-3(loại)
Bài 8:
a: \(\left|x^2-5x+4\right|=x+4\)
=>\(\begin{cases}x+4\ge0\\ x^2-5x+4=\left(x+4\right)^2\end{cases}\)
=>\(\begin{cases}x\ge-4\\ x^2-5x+4-x^2-8x-16=0\end{cases}\Rightarrow\begin{cases}x\ge-4\\ -13x-12=0\end{cases}\)
=>\(x=-\frac{12}{13}\)
b: \(\left|x^2-7x+12\right|=15-5x\)
=>\(\begin{cases}15-5x\ge0\\ \left(15-5x\right)^2=x^2-7x+12\end{cases}\)
=>\(\begin{cases}5x\le15\\ 25\left(x-3\right)^2=\left(x-3\right)\left(x-4\right)\end{cases}\Rightarrow\begin{cases}x\le3\\ \left(x-3\right)\left(x-4\right)-25\left(x-3\right)^2=0\end{cases}\)
=>\(\begin{cases}x\le3\\ \left(x-3\right)\left(x-4-25x+75\right)=0\end{cases}\Rightarrow\begin{cases}x\le3\\ \left(x-3\right)\left(-24x+71\right)=0\end{cases}\)
=>x=3(nhận) hoặc x=71/24(nhận)
c: \(\left|x^2-6x+5\right|+1=x\)
=>\(\left|x^2-6x+5\right|=x-1\)
=>\(\begin{cases}x-1\ge0\\ x^2-6x+5=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge1\\ x^2-6x+5=x^2-2x+1\end{cases}\)
=>\(\begin{cases}x\ge1\\ -6x+5=-2x+1\end{cases}\Rightarrow\begin{cases}x\ge1\\ -4x=-4\end{cases}\Rightarrow x=1\)
d: \(3x^2+5\left|x-3\right|+7=0\) (1)
TH1: x>=3
(1) sẽ trở thành: \(3x^2+5\left(x-3\right)+7=0\)
=>\(3x^2+5x-15+7=0\)
=>\(3x^2+5x-8=0\)
=>\(3x^2+8x-3x-8=0\)
=>(3x+8)(x-1)=0
=>x=-8/3(loại) hoặc x=1(loại)
TH2: x<3
(1) sẽ trở thành: \(3x^2+5\left(3-x\right)+7=0\)
=>\(3x^2+15-5x+7=0\)
=>\(3x^2-5x+22=0\)
\(\Delta=\left(-5\right)^2-4\cdot3\cdot22=25-12\cdot22<0\)
=>Phương trình vô nghiệm
e: ĐKXĐ: x<>2
\(\frac{x^2-1}{\left|x-2\right|}=x\)
=>\(x^2-1=x\cdot\left|x-2\right|\) (1)
TH1: x>2
(1) sẽ trở thành:
\(x\left(x-2\right)=x^2-1\)
=>\(x^2-2x=x^2-1\)
=>-2x=-1
=>x=1/2(loại)
TH2: x<2
(1) sẽ trở thành: \(x\left(x-2\right)=1-x^2\)
=>\(1-x^2=x^2-2x\)
=>\(x^2-2x+x^2-1=0\)
=>\(2x^2-2x-1=0\)
=>\(x^2-x-\frac12=0\)
=>\(x^2-x+\frac14-\frac34=0\)
=>\(\left(x-\frac12\right)^2=\frac34\)
=>\(\left[\begin{array}{l}x-\frac12=\frac{\sqrt3}{2}\\ x-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt3+1}{2}\left(nhận\right)\\ x=\frac{1-\sqrt3}{2}\left(nhận\right)\end{array}\right.\)
f: \(\frac{\left|x-1\right|}{x^2-x-6}=1\)
=>\(x^2-x-6=\left|x-1\right|\)
=>\(\begin{cases}x^2-x-6\ge0\\ \left(x^2-x-6\right)^2=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}\left(x-3\right)\left(x+2\right)\ge0\\ \left(x^2-x-6-x+1\right)\left(x^2-x-6+x-1\right)=0\end{cases}\)
=>(x>=3 hoặc x<=-2) và \(\left(x^2-2x-5\right)\left(x^2-7\right)=0\)
=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}x^2-2x+1-6=0\\ x^2-7=0\end{array}\right.\)
=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}\left(x-1\right)^2=6\\ x^2=7\end{array}\right.\)
=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}x-1=\sqrt6\\ x-1=-\sqrt6\\ x=\pm\sqrt7\end{array}\right.\)
=>(x>=3 hoặc x<=-2) và x\(\in\left\lbrace\sqrt6+1;-\sqrt6+1;\sqrt7;-\sqrt7\right\rbrace\)
=>\(x\in\left\lbrace\sqrt6+1;-\sqrt7\right\rbrace\)
Bài 6:
a: ĐKXĐ: \(x^2-4<>0\)
=>(x-2)(x+2)<>0
=>x∉{2;-2}
\(\frac{x^2-3x+5}{x^2-4}=-1\)
=>\(x^2-3x+5=-x^2+4\)
=>\(2x^2-3x+1=0\)
=>(x-1)(2x-1)=0
=>\(\left[\begin{array}{l}x-1=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=\frac12\left(nhận\right)\end{array}\right.\)
b: ĐKXĐ: x∉{2;-2/3}
\(\frac{2x+1}{3x+2}=\frac{x+1}{x-2}\)
=>(3x+2)(x+1)=(2x+1)(x-2)
=>\(3x^2+3x+2x+2=2x^2-4x+x-2\)
=>\(3x^2+5x+2-2x^2+3x+2=0\)
=>\(x^2+8x+4=0\)
=>\(x^2+8x+16-12=0\)
=>\(\left(x+4\right)^2=12\)
=>\(\left[\begin{array}{l}x+4=2\sqrt3\\ x+4=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3-4\left(nhận\right)\\ x=-2\sqrt3-4\left(nhận\right)\end{array}\right.\)
c: ĐKXĐ: x∉{2;-3}
\(1+\frac{2}{x-2}=\frac{10}{x+3}-\frac{50}{\left(2-x\right)\left(x+3\right)}\)
=>\(1+\frac{2}{x-2}=\frac{10}{x+3}+\frac{50}{\left(x-2\right)\left(x+3\right)}\)
=>
\(\dfrac{21}{36}-\left(-\dfrac{11}{30}\right)=\dfrac{7}{12}+\dfrac{11}{30}=\dfrac{7.5+11.2}{60}=\dfrac{57}{60}=\dfrac{19}{20}\\ ----\\\dfrac{-4}{8}+\left(-\dfrac{3}{10}\right)=\dfrac{-1}{2}-\dfrac{3}{10}=\dfrac{-1.5-3}{10}=\dfrac{-8}{10}=-\dfrac{4}{5}\\ ----\\ \dfrac{7}{12}-\left(-\dfrac{9}{20}\right)=\dfrac{7}{12}+\dfrac{9}{20}=\dfrac{7.5+9.3}{60}=\dfrac{62}{60}=\dfrac{31}{30}\\ ---\\ \dfrac{-2}{5}+\left(-\dfrac{11}{30}\right)=-\dfrac{2}{5}-\dfrac{11}{30}=\dfrac{-2.6-11}{30}=-\dfrac{29}{30}\)
a: -5/17<0<2/7
b: 11/10>1>9/14
=>-11/10<-9/14
c: 7/6>1>99/100
\(A=1-2+3-4+5-6+7-8+...+99-100\)
\(A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(A=\left(-1\right).50\)
\(A=-50\)
\(B=1+3-5-7+9+11-...-397-399\)
\(B=1-2+2-2+2-...+2-2-399\)
\(B=1-399\)
\(B=-398\)
\(C=1-2-3+4+5-6-7+...+97-98-99+100\)
\(C=-1+1-1+1-...-1+1\)
\(C=0\)
\(D=2^{2024}-2^{2023}-...-1\)
\(D=2^{2024}-\left(2^0+2^1+2^2+...2^{2023}\right)\)
\(D=2^{2024}-\left(\dfrac{2^{2024}-1}{2-1}\right)\)
\(D=2^{2024}-\left(2^{2024}-1\right)\)
\(D=2^{2024}-2^{2024}+1\)
\(D=1\)
A = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 +...+ 99 - 100
A = (1 - 2) + ( 3 - 4) + ( 5- 6) +....+(99 - 100)
Xét dãy số 1; 3; 5;...;99
Dãy số trên là dãy số cách đều có khoảng cách là: 3 - 1 = 2
Dãy số trên có số số hạng là: (99 - 1) : 2 + 1 = 50 (số)
Vậy tổng A có 50 nhóm, mỗi nhóm có giá trị là: 1- 2 = -1
A = - 1\(\times\)50 = -50
b,
B = 1 + 3 - 5 - 7 + 9 + 11-...- 397 - 399
B = ( 1 + 3 - 5 - 7) + ( 9 + 11 - 13 - 15) + ...+( 393 + 395 - 397 - 399)
B = -8 + (-8) +...+ (-8)
Xét dãy số 1; 9; ...;393
Dãy số trên là dãy số cách đều có khoảng cách là: 9-1 = 8
Dãy số trên có số số hạng là: ( 393 - 1): 8 + 1 = 50 (số hạng)
Tổng B có 50 nhóm mỗi nhóm có giá trị là -8
B = -8 \(\times\) 50 = - 400
c,
C = 1 - 2 - 3 + 4 + 5 - 6 +...+ 97 - 98 - 99 +100
C = ( 1 - 2 - 3 + 4) + ( 5 - 6 - 7+ 8) +...+ ( 97 - 98 - 99 + 100)
C = 0 + 0 + 0 +...+0
C = 0
d, D = 22024 - 22023- ... +2 - 1
2D = 22005- 22004 + 22003+...- 2
2D + D = 22005 - 1
3D = 22005 - 1
D = (22005 - 1): 3
a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
mik biết làm e hk hak
e, = 3,5-(-1):(-0,5)
= 3,5 - 2
= 1.5
1)\(\dfrac{2}{9}+\dfrac{-3}{4}+\dfrac{5}{30}\)
\(=\dfrac{2.20}{9.20}+\dfrac{-3.45}{4.45}+\dfrac{5.6}{30.6}\)
\(=\dfrac{40}{180}+\dfrac{-135}{180}+\dfrac{30}{180}\)
\(=\dfrac{40+\left(-135\right)+30}{180}\)
\(=\dfrac{-65}{180}\)
\(=\dfrac{-13}{36}\)
2)\(\dfrac{-7}{12}-\dfrac{11}{18}\)
\(=\dfrac{-7.3}{12.3}-\dfrac{11.2}{18.2}\)
\(=\dfrac{-21}{36}-\dfrac{22}{36}\)
\(=\dfrac{-21-22}{36}\)
\(=\dfrac{-43}{36}\)
3)\(\dfrac{7}{8}-\dfrac{-5}{16}\)
\(=\dfrac{7.2}{8.2}-\dfrac{-5}{16}\)
\(=\dfrac{14}{16}-\dfrac{-5}{16}\)
\(=\dfrac{14-\left(-5\right)}{16}\)
\(=\dfrac{19}{16}\)
4)\(\dfrac{3}{8}-\dfrac{-9}{10}-\dfrac{5}{16}\)
\(=\dfrac{3.10}{8.10}-\dfrac{-9.8}{10.8}-\dfrac{5.5}{16.5}\)
\(=\dfrac{30}{80}-\dfrac{-72}{80}-\dfrac{25}{80}\)
\(=\dfrac{30-\left(-72\right)-25}{80}\)
\(=\dfrac{77}{80}\)
-5/7 . 2/11 + (-5/7) . 9/11 + 5/7
= -5/7 . 2/11 + -5/7 . 9/11 + (-5/7) . (-1)
= (-5/7) . (2/11 + 9/11 -1)
= (-5/7) . 0
=0
ks nha bạn