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Câu 1a : tự kết luận nhé
\(2\left(x+3\right)=5x-4\Leftrightarrow2x+6=5x-4\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)
Câu 1b : \(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow x+3-2x+6=5-2x\Leftrightarrow-x+9=5-2x\Leftrightarrow x=-4\)
c, \(\frac{x+1}{2}\ge\frac{2x-2}{3}\Leftrightarrow\frac{x+1}{2}-\frac{2x-2}{3}\ge0\)
\(\Leftrightarrow\frac{3x+3-4x+8}{6}\ge0\Rightarrow-x+11\ge0\Leftrightarrow x\le11\)vì 6 >= 0
1) 2(x + 3) = 5x - 4
<=> 2x + 6 = 5x - 4
<=> 3x = 10
<=> x = 10/3
Vậy x = 10/3 là nghiệm phương trình
b) ĐKXĐ : \(x\ne\pm3\)
\(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)
=> \(\frac{x+3-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{5-2x}{\left(x-3\right)\left(x+3\right)}\)
=> x + 3 - 2(x - 3) = 5 - 2x
<=> -x + 9 = 5 - 2x
<=> x = -4 (tm)
Vậy x = -4 là nghiệm phương trình
c) \(\frac{x+1}{2}\ge\frac{2x-2}{3}\)
<=> \(6.\frac{x+1}{2}\ge6.\frac{2x-2}{3}\)
<=> 3(x + 1) \(\ge\)2(2x - 2)
<=> 3x + 3 \(\ge\)4x - 4
<=> 7 \(\ge\)x
<=> x \(\le7\)
Vậy x \(\le\)7 là nghiệm của bất phương trình
Biểu diễn
-----------------------|-----------]|-/-/-/-/-/-/>
0 7
a, Vì \(2+\frac{3-2x}{5}\)không nhỏ hơn \(\frac{x+3}{4}-x\)
\(\Rightarrow2+\frac{3-2x}{5}\ge\frac{x+3}{4}-x\)
Giải phương trình :
\(2+\frac{3-2x}{5}\ge\frac{x+3}{4}-x\)
\(\Rightarrow\frac{40}{20}+\frac{4\left(3-2x\right)}{20}\ge\frac{5\left(x-3\right)}{20}-\frac{20x}{20}\)
\(\Rightarrow40+12-8x\ge5x-15-20x\)
\(\Rightarrow7x=67\)
\(\Rightarrow x\ge\frac{67}{7}\)
b, \(\frac{2x+1}{6}-\frac{x-2}{9}>-3\)
\(\Rightarrow\frac{3\left(2x+1\right)}{18}-\frac{2\left(x-2\right)}{18}>\frac{-54}{18}\)
\(\Rightarrow6x+3-2x+4>-54\)
\(\Rightarrow4x>-61\)
\(\Rightarrow x>\frac{-61}{4}\)\(\left(1\right)\)
Và : \(x-\frac{x-3}{4}\ge3-\frac{x-3}{12}\)
\(\frac{12x}{12}-\frac{3\left(x-3\right)}{12}\ge\frac{36}{12}-\frac{x-3}{12}\)
\(\Rightarrow12x-3x+9\ge36-x+3\)
\(\Rightarrow10x\ge30\)
\(\Rightarrow x\ge3\)\(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\hept{\begin{cases}x>\frac{-61}{4}\\x\ge3\end{cases}\Rightarrow x>3}\)
Vậy với giá trị x > 3 thì x là nghiệm chung của cả 2 bất phương trình
A/ \(2\left(5x-3\right)=7x-18.\)
\(10x-6=7x-18\)
\(10-7x=6-18\)
\(3x=-12\)
\(x=-\frac{12}{3}=4\)
\(\Rightarrow S=\left\{4\right\}\)
B/ \(3x\left(x-2\right)+2x-4=0\)
\(3x\left(x-2\right)+2\left(x-2\right)=0\)
\(\left(x-2\right)\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x-2=0\Rightarrow x=2\\3x+2=0\Rightarrow3x=-2\Rightarrow x=-\frac{2}{3}\end{cases}}\)
\(\Rightarrow S=\left\{2;-\frac{2}{3}\right\}\)
C/ \(\frac{x+2}{3}\frac{x-3}{2}=\frac{x+5}{4}\)
\(\frac{\left(x+2\right)\left(x-3\right)}{3.2}=\frac{x+5}{4}\)
\(\frac{x^2-3x+2x-6}{6}=\frac{x+5}{4}\)
\(\frac{x^2-x-6}{6}=\frac{x+5}{4}\)
\(\frac{2\left(x^2-x-6\right)}{12}=\frac{3\left(x+5\right)}{12}\)
\(\frac{2x^2-2x-12}{12}=\frac{3x+15}{12}\)
\(\Rightarrow2x^2-2x-12=3x+15\)
(chuyển vế r làm tiếp)
Bài 1 :
\(a,2\left(5x-3\right)=7x-18\)
\(\Leftrightarrow10x-6=7x-18\)
\(\Leftrightarrow10x-7x=6-18\)
\(\Leftrightarrow3x=-12\)
\(\Leftrightarrow x=-4\)
PT có nghiệm S = { -4 }
\(b,3x\left(x-2\right)+2x-4=0\)
\(\Leftrightarrow3x^2-6x+2x-4=0\)
\(\Leftrightarrow3x^2-4x-4=0\)
\(\Leftrightarrow3x^2-6x+2x-4=0\)
\(\Leftrightarrow3x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+2=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=2\end{cases}}\)
KL : ............
\(c,\frac{x+2}{3}-\frac{x-3}{2}=\frac{x+5}{4}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}-\frac{6\left(x-3\right)}{12}=\frac{3\left(x+5\right)}{12}\)
\(\Leftrightarrow4x+8-6x+18=3x+15\)
\(\Leftrightarrow4x-6x-3x=-8-18+15\)
\(\Leftrightarrow x=-9\)
KL : .......
Ta có (a + b + c)2 \(\ge0\forall a;b;c\inℝ\)
=> a2 + b2 + c2 + 2ab + 2bc + 2ca \(\ge\)0
=> a2 + b2 + c2 \(\ge\)0 - (2ab + 2bc + 2ca)
=> a2 + b2 + c2 \(\le\)2ab + 2bc + 2ca
=> a2 + b2 + c2 \(\le\)2(ab + bc + ca)
Dấu "=" xảy ra <=> a + b + c = 0
Xí bài 2 ý a) trước :>
4x2 + 2y2 + 2z2 - 4xy - 4xz + 2yz - 6y - 10z + 34 = 0
<=> ( 4x2 - 4xy + y2 - 4xz + 2yz + z2 ) + ( y2 - 6y + 9 ) + ( z2 - 10z + 25 ) = 0
<=> [ ( 4x2 - 4xy + y2 ) - 2( 2x - y )z + z2 ] + ( y - 3 )2 + ( z - 5 )2 = 0
<=> [ ( 2x - y )2 - 2( 2x - y )z + z2 ] + ( y - 3 )2 + ( z - 5 )2 = 0
<=> ( 2x - y - z )2 + ( y - 3 )2 + ( z - 5 )2 = 0
Ta có : \(\hept{\begin{cases}\left(2x-y-z\right)^2\\\left(y-3\right)^2\\\left(z-5\right)^2\end{cases}}\ge0\forall x,y,z\Rightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\)
Thế vào T ta được :
\(T=\left(4-4\right)^{2014}+\left(3-4\right)^{2014}+\left(5-4\right)^{2014}\)
\(T=0+1+1=2\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
TL:
\(\Leftrightarrow\)\(\frac{15+3x+12}{15}< \frac{15x-5x-15}{15}\)
\(\Leftrightarrow27+3x< 10x-15\)
\(\Leftrightarrow7x>42\)
\(\Leftrightarrow x>6\)
Bài 7
\(a,A=x^2-2x+5\)
\(=\left(x^2-2x+1\right)+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
GTNN \(A=4\) khi \(\left(x-1\right)^2=0\Rightarrow x=1\)
\(b,B=x^2-x+1\)
\(=\left(x^2-2\cdot\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(c,C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x=t\)
\(\Rightarrow C=\left(t-6\right)\left(t+6\right)\)
\(=t^2-36\)
\(\left(x^2+5x\right)^2-36\ge36\forall x\)
\(d,D=x^2+5y^2-2xy+4y-3\)
\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)-4\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2-4\ge-4\)
Ta có (x - 2)2 - x2 - 8x +3 \(\ge0\)
<=> x2 - 4x + 4 - x2 - 8x + 3 \(\ge0\)
<=> - 12x + 7 \(\ge0\)
<=> -12x \(\ge-7\)
<=> \(x\le\frac{7}{12}\)
=> Số nguyên lớn nhất thỏa mãn bất phương trình là x = 0
\(\left(x-2\right)^2-x^2-8x+3\ge0\)
\(\Leftrightarrow x^2-4x+4-x^2-8x+3\ge0\)
\(\Leftrightarrow-12x+7\ge0\)
\(\Leftrightarrow x\le\frac{7}{12}\)
Vậy nghiệm của bất phương trình là \(x\le\frac{7}{12}\)
x + 4 5 − x + 5 < x + 3 3 − x − 2 2
Û 6(x + 4) - 30x + 150 < 10(x + 3) - 15(x - 2)
Û 6x + 24 - 30x + 150 < 10x + 30 - 15x + 30
Û 6x - 30x - 10x + 15x < 30 + 30 - 24 - 150
Û -19x < -114
Û x > 6
Vậy S = { x | x > 6 }
Nghiệm nguyên nhỏ nhất là x = 7.
Đáp án cần chọn là: A